Alternating Series test

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How to test Alternating Series

Any series with alternate signs is called alternating series  and can be represented as,

\mathbf{\sum_{n=1}^{\infty }(-1)^{n}a_{n}}

To test convergence of this type of series, we use alternating series test(AST). This test says:

For a given series   \mathbf{\sum_{n=1}^{\infty }(-1)^{n}a_{n}}  , where a_{n}\geq 0  for all n and if

  1. \mathbf{\lim_{n \to \infty }a_{n}=0},
  2.  If  \left \{ a_{n} \right \}   is decreasing sequence  i.e  a_{n+1}\leq a_{n}    for all  n, Then given series is convergent.

 

Example1 :Check the convergence of following alternating series.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }(-1)^{n-1}\frac{n^{2}}{n^{3}+1}}
Solution: Here we check the two conditions for alternating series test(AST).

  1. \dpi{120} \lim_{n \to \infty }a_{n}=\lim_{n \to \infty }\frac{n^{2}}{n^{3}+1}

\lim_{n \to \infty }\frac{n^{2}}{n^{2}(n+1)}=\lim_{n \to \infty }\frac{1}{n+1}

\dpi{120} \frac{1}{\infty +1}=\frac{1}{\infty }=0

2) Clearly this is a decreasing sequence because denominator is much larger than numerator and denominator  increase largely with increase  in value of n. Therefore terms of sequence  are decrease  as n increase.

Since both conditions of AST are met  so this series is convergent !

 

Example2: Determine if the following series is convergent or divergent.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }(-1)^{n}\frac{n^{2}}{n^{2}+3}}

Solution:  Because of alternating signs \dpi{120} (-1)^{n} , This is alternating series and we use AST  to check its convergence.

We check the two conditions one by one.

  1. \dpi{120} \lim_{n \to \infty }a_{n}=\lim_{n \to \infty }\frac{n^{2}}{n^{2}+3}

\lim_{n \to \infty }\frac{n^{2}}{n^{2}(1+\frac{3}{n^{2}})}=\lim_{n \to \infty }\frac{1}{1+\frac{3}{n^{2}}}

In above step we can divide  numerator and denominator by n^2 instead of factoring out.

\dpi{120} \frac{1}{1+\frac{3}{\infty }}=\frac{1}{1+0}=1

Here we got the limit  as 1 . First condition is not met so no need to check the second condition.

This series is not convergent .

 

Example3: Determine if the following series is convergent or divergent.

\dpi{120} {\color{Red} \sum_{n=2}^{\infty }\frac{cos(n\pi )}{\sqrt{n}}}

Solution:  You will be wondering how come this is an alternating series. Lets  see,

We know that

 cos(π) = -1 ,     cos(2π) =1,    cos(3π) = -1 ……….so on

so we can write   cos(nπ) = \dpi{120} (-1)^{n}

So the given series is actually an alternating series.

\dpi{120} \sum_{n=2}^{\infty }\frac{cos(n\pi )}{\sqrt{n}}=\sum_{n=2}^{\infty }\frac{(-1)^{n}}{\sqrt{n}}

Checking  the two conditions for AST

  1. \dpi{120} \lim_{n \to \infty }a_{n}=\lim_{n \to \infty }\frac{1}{\sqrt{n}}=\frac{1}{\infty }=0

2.  \dpi{120} a_{n}=\frac{1}{\sqrt{n}}  which is clearly a decreasing  sequence as its denominator get larger with increase in n.

Though it can also be checked using derivative test.

We know that function is decreasing when first derivative is negative.

\dpi{120} \left (a_{n} \right )' =\frac{-1}{2\sqrt{n}}<0

 

So this series is decreasing one.

Since both conditions are met so this series is Convergent by AST.

 

 

 

 

 

 

Practice problems:

Check the convergence of  following series :

1. \dpi{120} \sum_{n=1}^{\infty }(-1)^{n-1}\frac{1}{2n+1}

2. \dpi{120} \sum_{n=1}^{\infty }(-1)^{n}\frac{3n^{3}}{n^{3}+3}

3.  \dpi{120} \sum_{n=1}^{\infty }\frac{cos(n\pi )}{n^{2}}

 

 

 

 

 

Answers:

  • Convergent
  • Divergent
  • Convergent

 

 

Absolute and Conditional convergence

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How to check absolute convergence

Absolute convergence: If the absolute value \dpi{120} \sum \left | a_{n} \right |   of the series  , \dpi{120} \sum a_{n}  is convergent , then series  is called absolutely convergent.

Absolute convergence is a  stronger type of convergence.  Series that are absolutely convergent, are guaranteed to be convergent.  However, series that are convergent may or may not be absolutely convergent.

Lets look into some examples.

 

Example1: Does the following series converge absolutely.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n^{3}}}

Solution: lets check the absolute value of this series first.

\dpi{120} \sum_{n=1}^{\infty }\left | a_{n} \right |=\sum_{n=1}^{\infty }\left |\frac{(-1)^{n+1}}{n^{3}} \right |= \sum_{n=1}^{\infty }\frac{1}{n^{3}}

Which is a P series .

Here p=3  which is greater than 1

The series is  convergent using P series test. So the given series converges absolutely

 

Example2: Check the following series for absolute convergence.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }\frac{sin(n)}{n^{2}}}

Solution:  First note that this is not alternating series but we can check its absolute convergence.

We know that sin ranges from -1 to 1

-1 ≤  Sin(n) ≤ 1

Which can be written as,  |sin(n)| ≤  1

Therefore,      \dpi{120} \left | \frac{sin(n)}{n^{2}} \right |\leq \frac{1}{n^{2}}

We know that  \dpi{120} \frac{1}{n^{2}}   is convergent by P series test as p=2 >1 .

By comparison test  given series is convergent.

Therefore the given series is absolutely convergent.

 

Conditional convergence:

When given series \dpi{120} \sum a_{n} is  convergent but  \dpi{120} \sum \left |a_{n} \right |  is divergent then  we say series is conditionally convergent.

Let\dpi{120} \sum a_{n} is a convergent series. We say that this series is Absolutely Convergent if \dpi{120} \sum \left |a_{n} \right |  is also convergent. We say the original series is Conditionally Convergent  if  \dpi{120} \sum \left |a_{n} \right |  is not convergent.

 

Example3: Check whether the following series is absolutely convergent ,  conditionally convergent or divergent.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }\frac{(-1)^{n}}{n}}
Solution:  This is an alternating harmonic series  and we use AST test to check its convergence, which is covered in previous section. Using AST we check its two conditions.

  1. \dpi{120} \lim_{n \to \infty }a_{n}=\lim_{n \to \infty }\frac{1}{n}= \frac{1}{\infty }=0
  2. This is a decreasing series because terms get decreased when denominators  get increased, which can be checked using derivative test too.

\dpi{120} (a_{n})'=\left (\frac{1}{n} \right )' = \frac{-1}{n^{2}}    which is always negative , that means this function(series) is  always decreasing.

Since both conditions  are met so this series is convergent  by AST.

Now we check its absolute convergence.

\dpi{120} \sum_{n=1}^{\infty }\left | a_{n} \right |=\sum_{n=1}^{\infty }\left |\frac{(-1)^{n}}{n} \right |= \sum_{n=1}^{\infty }\frac{1}{n}

Which is divergent being a harmonic series.

Here we saw that original series  is convergent by AST  but  it is not convergent absolutely  so this series is conditionally convergent.

 

 

Example4: Check whether the following series is absolutely convergent ,  conditionally convergent or divergent.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }(-1)^{n}\left ( n+\frac{1}{n} \right )}
Solution:  First we check  its convergence using AST . Lets check the two conditions:

1)   \dpi{120} \lim_{n \to \infty }a_{n}=\lim_{n \to \infty }\left (n+\frac{1}{n} \right )= \infty +0 =\infty

First condition is not satisfied , no need to check it further.

This series  is divergent by AST

We can also check it  is not absolutely convergent  using Limit comparison test.

So given series is divergent.

 

 

 

 

 

 

 

Practice problems:

In the following exercises determine whether the series is absolutely convergent, conditionally convergent, or divergent.

  1. \dpi{120} \sum_{n=1}^{\infty }\frac{(-1)^{n}}{2n}
  2. \dpi{120} \sum_{n=1}^{\infty }(-1)^{n}\frac{n^{2}+1}{n^{2}}
  3. \dpi{120} \sum_{n=1}^{\infty }\frac{(-1)^{n}}{2^{n}}      [ geometric series ]

 

 

 

 

 

 

Answers:

  • Conditionally convergent
  • Divergent
  • Convergent

 

 

 

 

 

Volume of solid of revolution by three methods

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Volume of solid of revolution

Solid of revolution is generated by revolving a plane curve about  x or y axis or about any line that lie in the same plane as the given function. This line is called axis of revolution.

Volume of solids of revolution can be found using following three ways:

  1. Disk method
  2. Washer method(method of rings)
  3. Shell method

Lets discuss them one by one.

1) Disk method:

In this method when a curve y= f(x) defined and continuous on [a,b] is rotated  about  x or y axis, then a solid shape is obtained without any hole at the center. In this method cross section is taken perpendicular  to  axis of revolution.

When  region y=f(x)  is rotated about x axis:  Then cross  section is vertical (perpendicular to x axis) and  approximating   rectangle of cross section  has width dx.

Formula is given as…

\dpi{120} \mathbf{V=\int_{a}^{b}\pi (f(x))^{2}dx}

 

Example1: Find the volume of the solid obtained by revolving the region under the graph of y=√x  on [0,2] about the x-axis.

Solution: Here region R given by y=√x  is being rotated  about x axis and we get resulting solid shape  as in (b). Cross  section is shown with green slice(disk) which is perpendicular to axis of rotation(x axis)  having width dx.

So the approximating rectangle having width  move from x=0 to x=2. Applying the  values into the volume formula we get,

\dpi{120} V=\int_{0}^{2}\pi (\sqrt{x})^{2}dx

\dpi{120} V=\int_{0}^{2}\pi (x)dx

\dpi{120} V=\pi \left [ \frac{x^{2}}{2} \right ]_{0}^{2}

V= π[4/2-0]   =    2π  cubic units

When the region x=f(y) is rotated about y axis : Then  cross section is horizontal (perpendicular  to y axis) and approximating rectangle of cross section has width dy. Following formula is used .

\dpi{120} \mathbf{V=\int_{a}^{b}\pi (f(y))^{2}dy}

 

Example2: Find the volume of the solid obtained by revolving the region bounded by the graphs of y=x^3 ,  y=8  and x=0  about the y-axis.

Solution:  First we should sketch the region and should get the idea about its shape. When region y=x^3  is rotated about y axis, we get a parabolic solid.

Here cross section would be horizontal  slice of width dy which is perpendicular  to axis of rotation(y axis). Approximating rectangle of width dy will move from y=0 to y=8 which forms the limits of integration. Since cross section is horizontal so we should have function in terms of y. Changing y=x^3  in terms of y we  get    \dpi{120} x=y^{\frac{1}{3}} .

Using the formula given above we get the volume as,

\dpi{120} V=\int_{0}^{8}\pi \left (y^{\frac{1}{3}} \right )^{2}dy

\dpi{120} V=\pi \int_{0}^{8}y^{\frac{2}{3}}dy

\dpi{120} V=\pi \left [ \frac{3}{5}y^{\frac{5}{3}} \right ]_{0}^{8}

\dpi{120} V=\frac{3\pi }{5}\left [ 8^{\frac{5}{3}}-0 \right ]

\dpi{120} V= \frac{3\pi }{5}(32)=\frac{96\pi }{5}  cubic units

Lets work on an example where   solid is obtained by rotating  the region about any  line  other than coordinate axes.

 

Example 3:  Find the volume of the solid generated by revolving the region between the parabola x=y^2+1  and the line x=3 about the line x=3 .

Solution: Here the curve x=y^2+1   is a parabola with vertex (1,0) and  opening right. It is bounded on right side by vertical line x=3 which is also the axis of revolution. Lets get its sketch first.

Since  axis of revolution is a vertical line(x=3)  so cross section would be horizontal having an approximating disk of width dy.

Next important thing is to find radius of this circular disk.

R(y)= 3-(y^2+1 ) = 2-y^2

To find limits of integration we find  intersecting points of this parabolic curve with line x=3. So we set both equations equal and solve.

3=y^2+1

2 = y^2

±√2 = y

That means this circular disk (horz. Cross section) will move from -√2 to √2

So the volume is given as,

\dpi{120} V=\int_{-\sqrt{2}}^{\sqrt{2}}\pi \left (R(y) \right )^{2}dy

\dpi{120} V=\int_{-\sqrt{2}}^{\sqrt{2}}\pi \left( 2-y^{2} \right )^{2}dy\dpi{120} V=2\pi \left [ 4\sqrt{2}-\frac{(4\sqrt{2})^{3}}{3}+\frac{(\sqrt{2})^{5}}{5}-0 \right ]

\dpi{120} V=\int_{-\sqrt{2}}^{\sqrt{2}}\pi \left(4-4y^{2}+y^{4} \right )dy

 

Since area is symmetrical about x axis so we can use property of definite integrals.

\dpi{120} V=2\pi \int_{0}^{\sqrt{2}} \left(4-4y^{2}+y^{4} \right )dy

\dpi{120} V=2\pi \left [ 4y-\frac{4y^{3}}{3}+\frac{y^{5}}{5} \right ]_{0}^{\sqrt{2}}

\dpi{120} V=2\pi \left ( \frac{32\sqrt{2}}{15} \right )=\frac{64\sqrt{2}}{15}\pi  cubic units

 

 

 

 

2) Washer method (method of rings):

If we obtain the solid of revolution with a hole inside it, then we use washer method because the cross section is in form of a washer. Since the area of washer (ring) is given as  \dpi{120} R^{2}-r^{2}    so we are going to use this  concept to find volume of solid using this method.

In this case too, cross section would be perpendicular to axis of rotation. So when  region is revolved about x axis then width of cross sectional ring will be dx and hence the integral function would be in terms of x.

Here outer radius R= f(x) and inner radius  r = g(x)

 

Example4: Find the volume of the solid obtained by revolving the region bounded by y= √x  and y= x  about the x-axis.

Solution: First we need to  find the intersecting points of these curves and sketch it. To find intersecting points, we set both equations  equal and solve.

√x= x

x  = x^2        [ squaring both sides]

x-x^2 = 0

x(x-1) =  0

x=0 ,  x=1

Here outer radius  of washer(in green) is  R=√x  and inner radius  is  r=x. Using  these values into volume formula we get,

\dpi{120} V=\pi \int_{0}^{1}R^{2}-r^{2}dx

\dpi{120} V=\pi \int_{0}^{1} x-x^{2}dx

\dpi{120} V=\pi \left [ \frac{x^{2}}{2}-\frac{x^{3}}{3} \right ]_{0}^{1}

\dpi{120} V=\pi \left [\frac{1 }{2}-\frac{1}{3} \right ]= \frac{\pi }{6}   cubic units

 

Lets work on an example where region is revolved about a vertical line other than y axis and hole is created at the center when it is revolved.

Example5: Find the volume of the solid obtained by rotating the region   under the graph of y= 9-x^2  for 0≤x≤3   about the vertical axis x=-2.

Solution: Since a hole is created at the center while revolving the region so we use  washer method to find volume. Here axis of revolution is  vertical line x=-2 , therefore cross section perpendicular to line x=-2 would have  width dy which means  function as well as integration limits will be in terms of y.

y= 9-x^2     =>                   x^2 = 9-y     =>             x= √(9-y)

Since vertex of this parabola is at (0,9) so limits would be from y=0   to y=9.

One important point to be noted here is that, all distances (radii) are measured from axis of rotation.

Therefore,        Outer radius   R  = √(9-y)  +2

Inner radius   r = 2

\dpi{120} V=\pi \int_{0}^{9}R^{2}-r^{2}dy

\dpi{120} V=\pi \int_{0}^{9}\left ( \sqrt{9-y} \right )^{2}-2^{2} dy

\dpi{120} V=\pi \int_{0}^{9}9-y+4\sqrt{9-y}+4-4 dy

\dpi{120} V=\pi \int_{0}^{9}9-y+4\sqrt{9-y} dy

\dpi{120} V=\pi \left [ 9y-\frac{y^{2}}{2}-\frac{8}{3}(9-y)^{\frac{3}{2}} \right ]_{0}^{9}

\dpi{120} V=\pi \left [ \left (81-\frac{81}{2}-0 \right )-\left ( 0-\frac{8}{3} (9)^{\frac{3}{2}}\right ) \right ]

\dpi{120} V=\pi \left ( \frac{81}{2}+72 \right )=\frac{225}{2}\pi  cubic units

 

3) Shell method:

When method of disk and washer is difficult to use, then we use shell method. In this method  cross section is taken parallel to axis of rotation. As the name suggests, cross sections are taken in the form of cylindrical shells. Therefore we make use of the formula for volume of cylindrical shell , V =2π (radius)(height). Let’s work  on some examples.

When region is revolved about Y axis:  Then radius is taken as x  and height is taken as the given function defined  and continuous  on [a,b] and use the following formula:

\dpi{120} \mathbf{V= \int_{a}^{b}2\pi xf(x)dx}

 

 

Example6: The region under the graph of  \dpi{120} {\color{Red} y=-x^{3}+3x^{2}}   on interval [0,3] is revolved about the y-axis. Find the volume of the resulting solid.

Solution: We can’t use washer method for this problem, as it is not possible to convert given function in terms of y. So we opt for shell method in which cross section is taken parallel to axis  of rotation( y axis here).

Each cross section has  radius as ‘x’ which is the distance of cross section from y axis and  thickness as  dx. Height is given by \dpi{120} y=-x^{3}+3x^{2}    on interval [0,3]. Using formula we get ,

\dpi{120} V=\int_{0}^{3}2\pi x(-x^{3}+3x^{2})dx

\dpi{120} V=2\pi \int_{0}^{3} -x^{4}+3x^{3}dx

\dpi{120} V=2\pi \left [ \frac{-x^{5}}{5} +\frac{3x^{4}}{4}\right ]_{0}^{3}

\dpi{120} V=2\pi \left [ \frac{-3^{5}}{5}+\frac{3(3)^{4}}{4}-0 \right ]

\dpi{120} V=2\pi \left [ \frac{-243}{5}+\frac{243}{4} \right ]=\frac{243}{10}\pi   cubic units

 

When region is revolved about X axis : Then radius is taken as  y which is the distance from axis of rotation to the function, height is taken as function in terms of y and use the following formula

\dpi{120} \mathbf{V=\int_{a}^{b}2\pi yf(y)dy}

 

Example7: Find the volume of the solid obtained by rotating the region under   \dpi{120} {\color{Red} y=9-x^{2}}  over [0,3]  about the x-axis.

Solution: Using shell method , cross section is always parallel to axis of rotation( x axis here) so each cross section will have thickness as  ‘dy’ and radius as y. Therefore we need to convert given function in terms of y.

y=9-x^2    =>      x^2   =>  9-y            x=√(9-y)

Also we need to get  integration limits  in terms of  y . For that we can plugin end points of given interval [0,3] into y=9-x^2  and

get  new bounds as y=0 to y= 9.

Using radius =y and height  x=√(9-y)   into the formula given above we  get,

\dpi{120} V=\int_{0}^{9}2\pi *y*\sqrt{9-y} dy

Using U substitution ,

Let u = 9-y

9-u = y

-du = dy

\dpi{120} V=\int_{0}^{9}2\pi *(9-u)*\sqrt{u}(-du)

\dpi{120} V=2\pi \int_{0}^{9} 9u^{\frac{1}{2}}-u^{\frac{3}{2}}du

\dpi{120} V=2\pi \left [ \frac{9u^{\frac{3}{2}}}{\frac{3}{2}} -\frac{u^{\frac{5}{2}}}{\frac{5}{2}}\right ]_{0}^{9}

\dpi{120} V=2\pi \left [ 6u^{\frac{3}{2}}-\frac{2u^{\frac{5}{2}}}{5} \right ]_{0}^{9}

\dpi{120} V=2\pi \left [ 162-\frac{486 }{5}-0\right]=\frac{648\pi}{5}  cubic units

 

 

 

Practice problems:

  1. Find the volume of the solid of revolution obtained by revolving the region bounded by y=√x , the x-axis and the line x=9 about the x-axis.
  2. Find the volume of the solid generated by revolving the region enclosed by X=2/y , Y=2 , Y=6 and X=0 about the y-axis.
  3. Find the volume of the solid generated by revolving about the x-axis the region bounded by Y= 6X and Y=6X^2 .
  4. Find the volume of the solid generated when the region bounded by these three equations, is revolved about the y-axis , Y=X^2 , X=1 ,  Y=0 .

 

 

 

Answers:

1)  \dpi{120} \frac{81\pi }{2} cubic units

2)  4π/3 cubic units

3)  24π/5  cubic units                                           4)   π/2  cubic units

 

 

 

Volume of solids by Cross Sections

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How to find Volume of solid by Cross Sections

This method is  different than solid of revolution, yet similar  in some manner. In this method  we find volume of  figures  whose cross sections are shapes other than circles such as  squares, triangles, semicircles etc.

For this method we  need to be familiar with area formulas of different shapes as under.

Area of square    \dpi{120} \mathbf{A= (side)^{2}}

Area of triangle   \dpi{120} \mathbf{A= \frac{1}{2}(base)(height)}

Area of equilateral triangle \dpi{120} \mathbf{A= \frac{\sqrt{3}}{4}(side)^{2}}

Semicircle of radius x , \dpi{120} \mathbf{A= \frac{\pi }{2}(x)^{2}}

Area of rectangle A=L*W

Square with diagonal of length x,  \dpi{120} \mathbf{A=\frac{x^{2}}{2}}

If the cross section is perpendicular  to x axis, then its width would be dx and hence area would be a function of x. Then its volume on interval [a,b] is given as,

\dpi{120} \mathbf{V=\int_{a}^{b}A(x)dx}

If area of cross section is perpendicular to y axis then its width would be dy and hence area would be a function of y. Then its volume on [a,b] is given as,

\dpi{120} \mathbf{V=\int_{a}^{b}A(y)dy}

Use the following steps to find volume  by cross sections.

  • Draw the region of the base of the solid.
  • Draw a representative slice of the solid with the correct orientation and note the thickness as dx or
  • Draw the shape of the cross section.
  • Label the important measurements of the solid in terms of x or y looking at the base.
  • Write dV the volume of one representative slice using geometry formulas.
  • Write dV in terms of x or
  • Write the integral for the volume V, looking at the base to determine where the slices start and stop.

 

Example1: Find the volume of the solid whose base is bounded by the circle  \dpi{120} {\color{Red} x^{2}+y^{2}=4}  , the cross sections perpendicular to the x-axis are squares.

Solution: Since cross sections are perpendicular to x axis , therefore  function would be in terms of x. As clear from  its graph, integration limits  would be from x=-2 to x=2.

 

Side of square cross section would be from  bottom end

\dpi{120} -\sqrt{4-x^{2}}   to top end \dpi{120} \sqrt{4-x^{2}}  .

So the total length of side  = \dpi{120} \sqrt{4-x^{2}}-(-\sqrt{4-x^{2}})

Side = \dpi{120} 2\sqrt{4-x^{2}}

 

Volume  \dpi{120} v= \int_{-2}^{2}(side)^{2}dx

\dpi{120} V=\int_{-2}^{2}\left ( 2\sqrt{4-x^{2}} \right )^{2}dx

\dpi{120} V= \int_{-2}^{2}4(4-x^{2})dx

\dpi{120} V= \int_{-2}^{2}16-4x^{2}dx

\dpi{120} V= \left [16x-\frac{4x^{3}}{3} \right ]_{-2}^{2}

\dpi{120} V= \left [\left ( 32-\frac{32}{3} \right )-\left ( -32+\frac{32}{3} \right ) \right ]

\dpi{120} V= 64-\frac{64}{3}= \frac{128}{3}   cubic units

 

Example2 :  Find the volume of a solid whose base is bounded by the circle  \dpi{120} {\color{Red} x^{2}+y^{2}=4} , the cross sections perpendicular to the x-axis  are equilateral triangles.

Solution: Since the cross sections are perpendicular to x axis, so function would be in terms of x with bounds from -2 to 2.

Side of equilateral triangle would be same  as side of  square in previous problem .Ends of equilateral cross section  are spanned by circle,  bottom end being \dpi{120} -\sqrt{4-x^{2}}   and top end being \dpi{120} -\sqrt{4-x^{2}} .

Volume  \dpi{120} V=\int_{-2}^{2}\frac{\sqrt{3}(side)^{2}}{4}dx

\dpi{120} V=\int_{-2}^{2}\frac{\sqrt{3}(2\sqrt{4-x^{2}})^{2}}{4}dx

\dpi{120} V=\int_{-2}^{2}\frac{\sqrt{3}*4(4-x^{2})}{4}dx

\dpi{120} V=\int_{-2}^{2}\sqrt{3}(4-x^{2})dx

\dpi{120} V=\sqrt{3}\left [ 4x-\frac{x^{3}}{3} \right ]_{-2}^{2}

\dpi{120} V=\sqrt{3}\left [ \left ( 8-\frac{8}{3} \right )-\left ( -8+\frac{8}{3} \right ) \right ]

\dpi{120} V=\sqrt{3}\left ( \frac{16}{3}+\frac{16}{3} \right )=\frac{32\sqrt{3}}{3}  cubic units

 

Example3 : Find the volume of the solid whose base is bounded  by y = x + 1 and y =x^2-1, the cross sections perpendicular to the x-axis are rectangles of height 5.

Solution: Given cross sections are rectangles  perpendicular to x axis  whose base is bounded by y=x+1 and y=x^2 -1 and height given as 5.

First, we find points of intersection of these two curves which would be the  limits of integration. Setting both functions equal and solving,

x+1 = x^2 -1

0 =x^2-x-2

0 = (x-2)(x+1)

x = -1, 2

Base of rectangle =(x+1)-(x^2-1)

=x-x^2+2

Volume \dpi{120} V=\int_{-1}^{2}(base)(height)dx

\dpi{120} V=\int_{-1}^{2}(x-x^{2}+2)5 dx

\dpi{120} V= 5\left [ \frac{x^{2}}{2}-\frac{x^{3}}{3}+2x \right ]_{-1}^{2}

\dpi{120} V=5\left ( \frac{20}{6}+\frac{7}{6} \right )=\frac{45}{2}  cubic units

 

Example4: Find the volume if the cross sections perpendicular to the y-axis of a right triangle bounded by line   \dpi{120} {\color{Red} y=\frac{3}{4}x-3}  , x axis and y axis, are semicircles.

Solution: Since the cross sections are semicircles perpendicular to Y axis, so function would be in terms of y.

Distance of line from y axis would be the diameter of semicircle ,So radius would be half the distance of line from y axis.

First we find function in terms of y.

\dpi{120} y=\frac{3}{4}x-3

\dpi{120} \frac{4}{3}(y+3)=x

This is the distance of line from y axis, which forms the diameter of semicircular cross sections.

Radius of semicircular cross section   \dpi{120} r=\frac{1}{2}*\frac{4}{3}(y+3)=\frac{2}{3}y+2

volume  \dpi{120} V=\int_{-3}^{0}\left ( \frac{1}{2}\pi r^{2} \right )dy

\dpi{120} V=\frac{\pi }{2}\int_{-3}^{0}\left ( \frac{2}{3}y+2 \right )^{2}dy

\dpi{120} V=\frac{\pi }{2}\int_{-3}^{0}\left ( \frac{4}{9}y^{2}+\frac{8}{3}y+4 \right )dy

\dpi{120} V=\frac{\pi }{2}\left [ \frac{4}{27}y^{3}+\frac{4}{3}y^{2} +4y\right ]_{-3}^{0}

\dpi{120} V=\frac{\pi }{2}\left [ 0-\left ( \frac{4}{27}(-3)^{3}+\frac{4}{3}(-3)^{2}+4(-3) \right ) \right ]

\dpi{120} V=\frac{\pi }{2}\left [ 0-(-4+12-12) \right ]=2\pi  cubic units

 

 

 

Practice problems:

1) Find the volume of the solid whose base is the region inside the circle x 2 + y 2 = 9 if cross sections taken perpendicular to the y‐axis are squares.

2) Find the volume of the solid whose base is the region bounded by the lines x + 4 y = 4, x = 0, and y = 0, if the cross sections taken perpendicular to the x‐axis are semicircles.

Answers :

  • 144 cubic units
  • π/6  cubic units

 

Area of Bounded Regions

/by

How to find Area  of Bounded Regions

Integration has a large number of applications. Here we shall use integration for finding the areas of  bounded  regions. In order to find area of bounded  regions, we may use the following  algorithm.

Algorithm:

  • Make a rough sketch showing the area to be found.
  • Slice the area into horizontal or vertical strips as the case may be and consider the corresponding approximating rectangle.
  • Find the area of approximating rectangle. If strip is parallel to y axis then its width is taken as Δx or dx and if it is parallel to x axis then its width is taken as Δy or dy.
  • If limits are not given then find the limits within which the approximating rectangle can move. Suppose  it can move between  x=a  to x=b  or y=c to y=d then bounded area is given as..

\dpi{120} \mathbf{\int_{a}^{b}f(x) dx}                   OR                  \dpi{120} \mathbf{\int_{c}^{d}f(y) dy}

Area of bounded regions can be categorized in following three types.

  1. Area bounded by curve and x axis.
  2. Area bounded by curve and y axis.
  3. Area between two curves.

Lets discuss them one by one.

1.  Area bounded by curve and x axis :

This area lie between curve and x axis and is bounded by two vertical lines x=a and x=b which  form the limits of integration later.

In this case formula to find area of  bounded region is given as,

\dpi{120} \mathbf{A= \int_{a}^{b}f(x) dx}

 

Example1: Find the region bounded by curve y= 2x-x^2 and x axis.

Solution: Given function clearly  represents a parabola opening downward. We slice this region into vertical strips. To find intersecting points of this curve with x axis, we set  function =0 and solve for x.

2x-x^2  =0             =>          x(2-x)=0

X=0  , 2-x=0      =>            x=0 , x=2

We got the intersecting points as x =0,2. That means approximating  rectangle of width dx(in black) will move from x=0 to x=2.

Area A = \dpi{120} \int_{a}^{b}y dx

\dpi{120} A= \int_{0}^{2}2x-x^{2} dx

\dpi{120} A=\left [x^{2}-\frac{x^{3}}{3} \right ]_{0}^{2}

\dpi{120} A=\left [ 2^{2} -\frac{2^{3}}{3}-0\right ]

\dpi{120} A= 4-\frac{8}{3}= \frac{4}{3} sq. units

 

Note: Separate calculations has to be carried out for areas under the curve which are above x axis and  which are below x axis because  area under the x axis has  a negative value. Total area is always  the sum of absolute values of all these areas.

Lets work on an example to understand this method.

 

Example2 :  Find the area of  region bounded by curve y=2x-x^2  from x=0 to x=3.

Solution: This is same curve as used in above example ,but with extended limits. It is better to draw and  understand  the sketch first.

Here approximating rectangle with width dx moves from x=0 to x=2  above x axis and then move from x=2 to x=3  below x axis. So we calculate those areas above  and below x axis separately and finally add their absolute values.

\dpi{120} A=\int_{0}^{2}2x-x^{2}dx +\left | \int_{2}^{3} 2x-x^{2} dx\right |

\dpi{120} A=\left [x^{2}-\frac{x^{3}}{3} \right ]_{0}^{2}+\left |\left [x^{2}-\frac{x^{3}}{3} \right ]_{2}^{3} \right |

\dpi{120} A=\left [ 4 -\frac{8}{3}\right ]+ \left |\left (3^{2}-\frac{3^{3}}{3} \right )-\left (4-\frac{8}{3} \right ) \right |

\dpi{120} A=\frac{4}{3}+\left | (9-9)-\frac{4}{3} \right |

\dpi{120} A=\frac{4}{3}+|0-\frac{4}{3}|=\frac{4}{3}+\frac{4}{3}=\frac{8}{3} sq. units

 

 

2. Area bounded by curve and Y axis:

This area lie between curve and y axis bounded by two horizontal lines y=c and y=d which forms the limits of integration later. In this case approximating rectangle will have width as dy and move from y=c to y=d.

In this case formula to find area is given as..

\dpi{120} \mathbf{A=\int_{c}^{d}g(y)dy}

 

Example3: Find exact area of region bounded by  y= arctan(x), y axis  and  y=pi/4

Solution: Since it is easier to integrate  tan(y) instead of  arctanx so we  assume horizontal  approximate rectangle of width dy which move from y=0 to y= pi/4 and integrate the function with respect to y. Here a sketch will help you to understand it better.

Here the approximating rectangle (in black) with width dy will move in shaded (bounded) region from y=0 to y=pi/4 so area is given as,

\dpi{120} A=\int_{0}^{\frac{\pi }{4}}tany dy                  [ y= \dpi{120} tan^{-1}(x)  =>  tany=x]

\dpi{120} A= ln|secx|_{0}^{\frac{\pi }{4}}

A = ln |sec(π/4 )|- ln|sec(0)|

= ln(√2) – ln(1)

=  ln(√2 )-0    =    ln(√2 ) sq. units

 

Example4: Find area of the region bounded by  curve y^2 =2y-x and the y axis.

Solution: Lets rewrite and simplify this equation.

y^2=2y-x

x = 2y -y^2

Completing square we get,

x  -1 = -y^2 +2y-1

(x-1) = -(y-1)^2  which  represents a parabola with vertex  at (1,1) and opening to left.

To get intersecting points we  set  2y – y^2 =0  and get   y=0,2 as  intersecting points with y axis.

In the sketch shown above , approximating rectangle of width dy, is horizontal and  move from y=0 to y=2

Area    \dpi{120} A=\int_{0}^{2} x dy

\dpi{120} A=\int_{0}^{2}2y-y^{2} dy

\dpi{120} A=\left [ y^{2} - \frac{y^{3}}{3}\right ]_{0}^{2}

\dpi{120} A= 4-\frac{8}{3}= \frac{4}{3}   sq. units

 

3. Area bounded between two curves.

There can be two different  possible cases.

  1. When area is given between two curves y=f(x) and y=g(x) such that f(x)>g(x) on an interval [a,b]

Then the formula for area is given as…

\dpi{120} \mathbf{A=\int_{a}^{b}\left | f(x)-g(x) \right |dx}            OR

\dpi{120} \mathbf{A=\int_{a}^{b}\left |upper function-lower function \right |dx}

Here  approximating  vertical rectangle  of width dx move from x=a to x=b.

 

2.  When area is given between two  curves x=f(y) and x=g(y) on an interval [c,d] such that f(y) > g(y).

In this case approximating horizontal rectangle of width dy move from y=c to y=d.

And the  formula for area is given as.

\dpi{120} \mathbf{A=\int_{c}^{d}\left | f(y)-g(y) \right |dy}                   OR

\dpi{120} \mathbf{A=\int_{c}^{d}\left | right function- left function \right |dy}

Lets  work on some examples  to understand it better.

 

 

Example5:   Find the area of region bounded by curves y=x^2  and  x=y^2

Solution:  y=x^2  is a parabola opening up and x=y^2  is also a parabola  opening right. Area between the curves would be sliced into vertical  strips .Lets look at its sketch.

To find intersecting points of these curves , we  use substitution and  solve for x. We substitute y with x^2.

\dpi{120} x=(x^{2})^{2}

\dpi{120} x-x^{4}=0

x(1-x^3)  = 0

x=0     or    1-x^3 =0

x =0   or        1=x^3    =>   x=0,1

So the approximating  vertical rectangle of width dx will move from x=0 to x=1 and area is given as,

\dpi{120} A=\int_{0}^{1}\left |upper function-lower function \right |dx

Here upper curve is x=y^2  which is changed to y=√x  as a function of x.

\dpi{120} A=\int_{0}^{1}\sqrt{x}-x^{2}dx

\dpi{120} A=\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{3}}{3} \right ]_{0}^{1}

\dpi{120} A=\left [ \frac{2x^{\frac{3}{2}}}{3}-\frac{x^{3}}{3} \right ]_{0}^{1}

\dpi{120} A=\left [ \frac{2}{3}-\frac{1}{3} \right ]= \frac{1}{3}    sq. units

 

Example6: Determine the area of region bounded by  x=-y^2 +10 and x=(y-2)^2

Solution: lets first sketch the region  and find intersecting points. For that we  set both equation equal and solve for x.

\dpi{120} -y^{2}+10=(y-2)^{2}

\dpi{120} -y^{2}+10=y^{2}-4y+4

\dpi{120} 0=2y^{2}-4y-6

0 =2(y-3)(y+1)

Y= -1 ,3

That means approximating horizontal  rectangle of width dy will move  from y=-1 to y=3. Since  curves are  functions of y so we use formula,

\dpi{120} A=\int_{-1}^{3}\left | right function- left function \right |dy

\dpi{120} A=\int_{-1}^{3}(-y^{2}+10)-(y-2)^{2}dy

\dpi{120} A=\int_{-1}^{3}-2y^{2}+4y+6 dy

\dpi{120} A=\left [ \frac{-2y^{3}}{3}+2y^{2} +6y\right ]_{-1}^{3} = \frac{64}{3}  sq. units

 

 

 

Practice problems:

  1. Find the area of shaded region bounded by parabola y=x^2  and the line y= x+2.
  2. Find the area of region bounded by x =4-y^2  and y axis.
  3. Find the area between the curve y=cosx and x axis  on the interval [0,π].
  4. Find the area between lines y=x and y=1 and curve y =x^2 /4  in the first quadrant.

 

 

 

 

 

 

Answers:

  • 9/2 units
  • 32/3 sq. units
  • 2  sq.units
  • 5/6 sq. units

Integration by Partial fractions

/by

How to do partial fraction decomposition.

Any proper rational function  \dpi{120} \frac{f(x)}{g(x)}   can be expressed as a sum of rational functions. Each such fraction is called partial fraction and the process of obtaining them is called decomposition of  into partial fractions.

The decomposition of   \dpi{120} \frac{f(x)}{g(x)}   into partial fractions mainly depand upon the nature of factors of g(x).

Case 1:

When denominator g(x) is expressible as the product of non – repeating  linear factors (x-a1)(x-a2)….(x-an) then rational function   \dpi{120} \frac{f(x)}{g(x)}  is decomposed  as..

\dpi{120} \mathbf{\frac{f(x)}{g(x)}= \frac{A_{1}}{(x-a_{1})}+\frac{A_{2}}{(x-a_{2})}+......+\frac{A_{n}}{(x-a_{n})}}

Where A1, A2,……An  are constants and can be determined by equating the numerators on left side and right side and then substituting  x=a1 , a2,…….. an

 

Example1:Resolve     \dpi{120} {\color{Red} \frac{3x+2}{x^{3}-6x^{2}+11x-6}}       into partial fractions.         

Solution: First we get factors of denominator as (x-1)(x-)(x-3). Since all are non repeated linear factors, so expression get decomposed like this:

\dpi{120} \frac{3x+2}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}

Multiply each term on both sides with (x-1)(x-2)(x-3) and we get, (This is short way of doing common denominators)

(3x+2) = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)

Values of constants A,B and C can be obtained by two ways.

First method:  by substituting zeros of denominator  as x values. Here zeros are 1,2 and 3 so we plugin x=1,2 and 3 one by one.

Let x=1 ,

(3*1+2) = A(1-2)(1-3)+B(1-1)(1-3)+C(1-1)(1-2)

5  =  A(-1)(-2) +B(0)(-2) +C(0)(-1)

5  =  2A

5/2 = A

Let x=2 ,

(3*2+2) = 0 +B(2-1)(2-3) +0

8  =  -B

-8 = B

Let x= 3,

(3*3+2) =  0+0+C(3-1)(3-2)

11 = 2C

11/2 = C

Using values of A,B and C given expression can be decomposed as,

\dpi{120} \frac{3x+2}{(x-1)(x-2)(x-3)}=\frac{5}{2(x-1)}-\frac{8}{x-2}+\frac{11}{2(x-3)}

 

Second method:By comparing coefficients of like terms. After getting rid of denominators on both sides, we multiply and simplify the terms on right side  like this..

(3x+2)= A(x-2)(x-3) +B(x-1)(x-3) +c(x-1)(x-2)

(3x+2) = A(x^2 -5x+6 )+B(x^2 -4x+3 )+C(x^2-3x+2 )

(3x+2) =x^2 (A+B+C) + x(-5A-4B-3C)+ 6A+3B+2C

Comparing coefficients of like terms we get following  three linear equations.

0=A+B+C                     3=-5A-4B-3C              2=6A+3B+2C

Solving these three linear equations by substitution or elimination we get A=5/2 ,B=-8 and C=11/2

For non repeated linear factors ,first method is always helpful while second method is helpful for all other types.

 

Case 2:

When denominator is expressible as the product of linear factors such that some of them are repeating.

\dpi{120} g(x)=(x-a)^{k}(x-a_{1})(x-a_{2})

then,

\dpi{120} \mathbf{\frac{f(x)}{g(x)}= \left [\frac{A_{1}}{(x-a)}+\frac{A_{2}}{(x-a)}+......+\frac{A_{k}}{(x-a)^{k}} \right ]+\frac{B_{1}}{(x-a_{1})}+\frac{B_{2}}{(x-a_{2})}}

[for repeated linear factors]                                      [non repeated]

 

Example2:Resolve  the given expression into partial fractions and then evaluate the integral.

\dpi{120} {\color{Red} \int \frac{x^{2}}{(x-1)^{3}(x+1)}dx}

Solution:

\dpi{120} \int \frac{x^{2}}{(x-1)^{3}(x+1)}dx= \frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}}+\frac{D}{(x+1)}

Multiply both sides with  \dpi{120} (x-1)^{3}(x+1)      we get,

\dpi{120} x^{2}=A(x-1)^{2}(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^{3}

Let x=1

1= A(0)(1)+B(0)(1)+C(1+1)+D(0)

1= 2C

½ = C

Let x=-1

(-1)^2= A(1)(0)+B(-2)(0)+C(0)+D

1  = -8D

-1/8  = D

Let x=0

0 =A(1)(1)+B(-1)(1)+C(1)+D(-1)

0 = A-B+C-D   ———-(i)

Let x = 2

4 = 3A+3B+3C+D  ——–(ii)

Plugin  C=1/2 and D=-1/8 in equations (i) and (ii) we get,

A-B =-5/8  and  A+B=7/8

Solving these two equations using elimination we get,

A = 1/8 , B= 3/4

Using values of A,B,C and D we get the expression decomposed  as,

\dpi{120} \frac{x^{2}}{(x-1)^{3}(x+1)}= \frac{1}{8(x-1)}+\frac{3}{4(x-1)^{2}}+\frac{1}{2(x-1)^{3}}-\frac{1}{8(x+1)}

And the integral becomes ,

\dpi{120} \int \frac{x^{2}}{(x-1)^{3}(x+1)}dx= \frac{1}{8}\int \frac{1}{(x-1)}dx+\frac{3}{4}\int \frac{1}{(x-1)^{2}}dx+\frac{1}{2}\int \frac{1}{(x-1)^{3}}dx \dpi{120} -\frac{1}{8}\int \frac{1}{(x+1)}dx

\dpi{120} \int \frac{x^{2}}{(x-1)^{3}(x+1)}dx= \frac{1}{8}ln|x-1|+\frac{3}{4}\frac{(x-1)^{-1}}{-1}+\frac{1}{2}\frac{(x-1)^{-2}}{-2} \dpi{120} -\frac{1}{8}ln|x+1|+c

\dpi{120} \int \frac{x^{2}}{(x-1)^{3}(x+1)}dx= \frac{1}{8}ln|x-1|-\frac{1}{8}ln|x+1|-\frac{3}{4(x-1)}-\frac{1}{4(x-1)^{2}}+c

\dpi{120} \int \frac{x^{2}}{(x-1)^{3}(x+1)}dx= \frac{1}{8}ln \left |\frac{x-1}{x+1} \right | -\frac{3}{4(x-1)}-\frac{1}{4(x-1)^{2}}+c

 

Case 3:

When some factors are quadratic along with some linear factors. Then for each linear factor we use numerator as constant alone  and for each quadratic factor we use numerator as linear function of the form Ax+B.

Lets work on a similar example to understand this case .

 

Example3: Resolve the given expression into partial fractions.

\dpi{120} {\color{Red}\frac{2x-1}{(x+1)(x^{2}+2)}dx}

Solution:

\dpi{120} \frac{2x-1}{(x+1)(x^{2}+2)}dx=\frac{A}{(x+1)}+\frac{Bx+C}{x^{2}+2}

Multiply both sides with (x+1)(x^2+2)

(2x-1) = A(x^2 +2)  +(Bx+C)(x+1)

(2x-1) = Ax^2 +2A+Bx^2 +(B+C)x + C

2x-1=  (A+B)x^2 +(B+C)x +2A+C

Comparing coefficients of like terms,

0=A+B                2=B+C                -1=2A+C

-A=B                    2=-A+C               -1=2A+C

Solving equations  2=-A+C and -1=2A+C together by subtracting

We get,                       A=-1

Using A=-1 we get   B=-(-1)= 1

Using A=-1 we get C=2+A=2-1=1

Using  A=-1 ,B=1 ,C=1  we get the given expression resolved as,

\dpi{120} \frac{2x-1}{(x+1)(x^{2}+2)}dx=\frac{-1}{(x+1)}+\frac{x+1}{x^{2}+2}

 

Important: If a rational function contains only even powers of x in both numerator and denominator , then x^2 can be replaced with some other variable (say t) and proceed same way using linear factors.

Example4 : Resolve into partial fractions  and then integrate the given expression.

\dpi{120} {\color{Red} \int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx}

Solution:  Since top and bottom functions both have even powers of x so we assume x^2= t

\dpi{120} \frac{x^{2}}{(x^{2}+1)(x^{2}+4)} =\frac{t}{(t+1)(t+4)}=\frac{A}{t+1}+\frac{B}{t+4}

t =  A(t+4) + B(t+1)

plugin t=-1, we get

-1 = A(-1+4) +B(0)

-1=3A

-1/3 = A

Plugin t=-4, we get

-4= A(0) +B(-3)

4/3 =B

Using values of A and B we get the  expression decomposed as..

\dpi{120} \frac{A}{t+1}+\frac{B}{t+4}=\frac{-1}{3(t+1)}+\frac{4}{3(t+4)}=\frac{-1}{3(x^{2}+1)}+\frac{4}{3(x^{2}+4)}

Taking integral,

\dpi{120} \int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx =\int \frac{-1}{3(x^{2}+1)}dx+\int \frac{4}{3(x^{2}+4)}dx

\dpi{120} \int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx =\frac{-1}{3}\int \frac{1}{(x^{2}+1)}dx+\frac{4}{3}\int \frac{1}{(x^{2}+4)}dx

\dpi{120} \int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx =\frac{-1}{3}tan^{-1}(x)+\frac{4}{3}*\frac{1}{2}tan^{-1}(\frac{x}{2})+c

\dpi{120} \int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx =\frac{-1}{3}tan^{-1}(x)+\frac{2}{3}tan^{-1}(\frac{x}{2})+c

 

 

Case 4:

When factors are quadratic and repeating.

Following example illustrates the procedure.

Example5: Resolve the given expression into partial fractions.

\dpi{120} {\color{Red} \int \frac{2x-3}{(x-1)(x^{2}+1)^{2}}dx}

Solution:

\dpi{120} \frac{2x-3}{(x-1)(x^{2}+1)^{2}}=\frac{A}{x-1}+\frac{Bx+c}{(x^{2}+1)}+\frac{Dx+E}{(x^{2}+1)^{2}}

Multiplying both sides with (x-1)  we get,

\dpi{120} 2x-3 =A(x^{2}+1)^{2}+(Bx+C)(x-1)(x^{2}+1)+(Dx+E)(x-1)

Plugin x=1 we get,

-1 = A(1+1)^2

-1 = 4A

-1/4 =A

Comparing coefficients of like terms we get,

A+B=0       C-B=0        2A+B-C+D=0    C+E-B-D=2        A-C-E=-3

Substitute A=-1/4  and solving these equations, we get

B=C=1/4            D=1/2 and   E= 5/2

Using values of A,B,C,D and E we get the given expression decomposed as,

\dpi{120} \frac{2x-3}{(x-1)(x^{2}+1)^{2}}=\frac{\frac{-1}{4}}{x-1}+\frac{\frac{1}{4}(x+1)}{(x^{2}+1)}+\frac{\frac{1}{2}x+\frac{5}{2}}{(x^{2}+1)^{2}}

\dpi{120} \frac{2x-3}{(x-1)(x^{2}+1)^{2}}=\frac{-1}{4(x-1)}+\frac{(x+1)}{4(x^{2}+1)}+\frac{x+5}{2(x^{2}+1)^{2}}

 

Here is an useful summary chart on partial decomposition rules.

 

Type of factors

In denominator

 Expression       Form of partial fractions
Linear factors

 

 

Repeated  linear factors

 

 

Irreducible Quadratic factors

 

 

Repeated quadratic factors

 

   \dpi{120} \mathbf{\frac{f(x)}{(x+a)(x+b)}}

 

\dpi{120} \mathbf{\frac{f(x)}{(x+a)^{k}}}

 

 

   \dpi{120} \mathbf{\frac{f(x)}{(x+a)(ax^{2}+bx+c)}}

    

        

 \dpi{120} \mathbf{\frac{f(x)}{(ax^{2}+bx+c)^{k}}}

 

 

 

         \dpi{120} \mathbf{\frac{A}{x+a}+\frac{B}{x+b}}

\dpi{120} \mathbf{\frac{A_{1}}{(x+a)}+\frac{A_{2}}{(x+a)^{2}}+.......+\frac{A_{n}}{(x+a)^{k}}}

 

 

\dpi{120} \mathbf{\frac{A}{(x+a)}+\frac{Bx+C}{(ax^{2}+bx+c)}}

 

 

\dpi{120} \mathbf{\frac{A_{1}x+B_{1}}{(ax^{2}+bx+c)}+\frac{A_{2}x+B_{2}}{(ax^{2}+bx+c)^{2}}+....+\frac{A_{k}x+B_{k}}{(ax^{2}+bx+c)^{k}}}

 

+

 

 

Practice problems:

Resolve the following expressions into partial fractions.

  1. \dpi{120} \frac{3x-2}{(x-1)^{2}(x+1)(x+2)}
  2. \dpi{120} \frac{8}{(x+2)(x^{2}+4)}

Evaluate the following integrals using partial fractions.

3. \dpi{120} \int \frac{2x+1}{(x+1)(x-2)}dx

4.\dpi{120} \int \frac{2x}{(x^{2}+1)(x^{2}+3)}dx

 

 

 

 

 

 

Answers:

1.   \dpi{120} \frac{13}{36(x-1)}+\frac{1}{6(x-1)^{2}}-\frac{5}{4(x+1)}+\frac{8}{9(x+2)}

2.    \dpi{120} \frac{1}{(x+2)}+\frac{-x+2}{x^{2}+4}

3.   \dpi{120} \frac{1}{3}ln|x+1|+\frac{5}{3}ln|x-2|+c

4.   \dpi{120} \frac{1}{2}ln\left | \frac{x^{2}+1}{x^{2}+3} \right |+c

 

 

 

 

Integration by Parts( IBP)

/by

How to solve  integral by parts .

This method is used for the integral of product of two different functions using the following formula.

\dpi{150} \mathbf{\int udv = uv-\int vdu}

Where u is assumed the first function and dv as the second function. Here we  should choose u and dv wisely, otherwise we may land in a complex situation. We should choose the function ‘u’ which comes first in the word ILATE, where

I :Inverse trigonometric functions

          L: Logarithmic functions

          A: Algebraic functions

          T: Trigonometric functions

          E: Exponential functions

Lets work on some examples  to understand this process.

 

Example1: \dpi{120} {\color{Red} \int x sinx dx}

Solution:   This  function is product of algebraic  and trigonometric function. Out of these two, algebraic function comes first in order of ILATE  so we assume x as u and sinx as dv.

Let u= x                       dv=sinx

du = dx                         v=ʃsinx dx  = -cosx

Now applying  the formula

\dpi{120} \int udv = uv-\int vdu

∫x sinx dx    =  x(-cosx) –∫-cosx dx

=  – xcosx  +∫cosx dx

=  -x cosx + sinx + c         [ answer]

 

Example2: \dpi{120} {\color{Red} \int x^{2}e^{x}dx} 

Solution:  sometimes  we may have to use  integration by parts process more than once. As per ILATE rule algebraic function comes first so we assume x^2 as u.

Let  x^2 = u                                    dv=e^x

2x dx= du                                 v = e^x

Applying  the formula,

\dpi{120} \int udv = uv-\int vdu

\dpi{120} \int x^{2}e^{x}dx =x^{2}e^{x}-\int 2xe^{x}dx

\dpi{120} \int x^{2}e^{x}dx= x^{2}e^{x}-2\int xe^{x}dx

Again applying integration by parts  for integral term,

We assume  u= x                      and  dv=e^x

du = dx                                             v= e^x

\dpi{120} \int x^{2}e^{x}dx =x^{2}e^{x}-2\left [ xe^{x}-\int e^{x}dx \right ]

\dpi{120} \int x^{2}e^{x}dx= x^{2}e^{x}-2xe^{x}+2e^{x}+c

 

 

Example3:  ∫ lnx dx  

Solution: To find integral of such functions  where  integrand is not the product of two functions , we consider another function as 1(constant), assume it as dv  and proceed  same way.

∫lnx dx         =  ∫ lnx *1 dx

Let  u= lnx                            dv =  1

du = 1/x                            v = x

Applying  formula,

\dpi{120} \int udv = uv-\int vdu

∫lnx*1 dx = x lnx -∫1/x *x dx

=  x lnx  – ∫1 dx

=  x lnx  – x + c                        [answer]

For integration of  other inverse functions like

\dpi{120} sin^{-1}(x), cos^{-1}(x), tan^{-1}(x)

etc. we use  same process.

 

Example4: \dpi{120} {\color{Red} \int sec^{3}(x) dx}

Solution:  Sometimes  for integration of single functions we have to break them up as product of two functions and continue with integration by parts.

\dpi{120} \int sec^{3}xdx=\int sec^{2}x secx dx

Both  functions being trigonometric, it is better to assume sec^2(x) as dv because  its integral is easy to find.

Let  u = secx                             dv = sec^2(x)

du = secx tanx                     v =  tanx

Applying  formula

\dpi{120} \int udv = uv-\int vdu

\dpi{120} \int sec^{3}xdx= secxtanx-\int secx tanx tanx dx

\dpi{120} \int sec^{3}xdx= secxtanx-\int secx tan^{2}xdx

\dpi{120} \int sec^{3}xdx= secxtanx-\int secx (sec^{2}x-1)dx

\dpi{120} \int sec^{3}xdx= secxtanx-\int sec^{3}xdx-\int secxdx

\dpi{120} \int sec^{3}xdx+\int sec^{3}xdx = secxtanx-\int secxdx

\dpi{120} 2\int sec^{3}xdx = secxtanx-ln|secx +tanx| +c

\dpi{120} \int sec^{3}xdx = \frac{1}{2}secxtanx-\frac{1}{2}ln|secx +tanx| +c

 

 

Example5:\dpi{120} {\color{Red} \int e^{2x}sin3xdx}

Solution: For such type of integrals we can ignore ILATE rule and assume any of the function as u. Answer will remain same irrespective of choice of u and dv.

Let    u =e^2x                                dv=  sin(3x)

du = 2e^2x                                      v = -cos(3x)/3

Applying  formula

\dpi{120} \int udv = uv-\int vdu

\dpi{120} \int e^{2x}sin3x dx=e^{2x}(\frac{-cos3x}{3})-\int 2e^{2x}(\frac{-cos3x}{3})dx

\dpi{120} \int e^{2x}sin3x dx=\frac{-1}{3}e^{2x}cos3x+\int \frac{2}{3}e^{2x}cos3xdx

Again applying formula for later part,

Let    u =e^2x                                   dv=  cos(3x)

du = 2e^2x                                         v =  sin(3x)/3

\dpi{120} \int e^{2x}sin3x dx=\frac{-1}{3}e^{2x}cos3x+\frac{2}{3}\left [ e^{2x}(\frac{sin3x}{3})-\int 2e^{2x}(\frac{sin3x}{3})dx \right ]

\dpi{120} \int e^{2x}sin3x dx=\frac{-1}{3}e^{2x}cos3x+\frac{2}{9} e^{2x}sin3x-\frac{4}{9}\int e^{2x}sin3xdx

\dpi{120} \int e^{2x}sin3x dx+\frac{4}{9}\int e^{2x}sin3xdx =\frac{-1}{3}e^{2x}cos3x+\frac{2}{9} e^{2x}sin3x

\dpi{120} \left ( 1+\frac{4}{9} \right )\int e^{2x}sin3x dx =\frac{1}{9}e^{2x}\left [-3cos3x+ 2sin3x \right ] +c

\dpi{120} \frac{13}{9}\int e^{2x}sin3x dx =\frac{1}{9}e^{2x}\left [-3cos3x+ 2sin3x \right ] +c

\dpi{120} \int e^{2x}sin3x dx =\frac{1}{13}e^{2x}\left [-3cos3x+ 2sin3x \right ]+c

 

Integral of the form \dpi{120} {\color{Red} \mathbf{\int e^{x}\left [ f(x)+f'(x) \right ]dx}}

Formula :   \dpi{150} \mathbf{\int e^{x}\left [ f(x)+f'(x) \right ]dx=e^{x}f(x)+c}

                

Example6 :\dpi{120} {\color{Red} \int e^{x}\left ( \frac{1}{x}-\frac{1}{x^{2}}\right )dx}

Solution:  First we simplify this expression,

\dpi{120} \int e^{x}\left ( \frac{1}{x}-\frac{1}{x^{2}}\right )dx=\int \frac{1}{x}e^{x}dx-\int \frac{1}{x^{2}}e^{x}dx

Solving only  first integral  and keeping second one same,

Let u= 1/x                       dv =e^x

\dpi{120} du= \frac{-1}{x^{2}}dx                   v =e^x

\dpi{120} \int e^{x}\left ( \frac{1}{x}-\frac{1}{x^{2}}\right )dx=\frac{1}{x}e^{x}-\int \frac{-1}{x^{2}}e^{x}dx-\int \frac{1}{x^{2}}e^{x}dx

\dpi{120} \int e^{x}\left ( \frac{1}{x}-\frac{1}{x^{2}}\right )dx=\frac{1}{x}e^{x}+\int \frac{1}{x^{2}}e^{x}dx-\int \frac{1}{x^{2}}e^{x}dx

\dpi{120} \int e^{x}\left ( \frac{1}{x}-\frac{1}{x^{2}}\right )dx=\frac{1}{x}e^{x}+c

This  answer can be verified  using formula given above. Here we were given f(x) =1/x and derivative f’(x)= -1/x^2 .

 

Example7:\dpi{120} {\color{Red} \int e^{x}\left ( tanx +lnsecx \right )dx}

Solution: First we simplify the expression and get,

\dpi{120} \int e^{x}\left ( tanx +lnsecx \right )dx=\int e^{x}ln secx dx+\int e^{x} tanxdx

Solving only first integral and keeping second one same,

Let u= ln secx                                         dv = e^x

\dpi{120} du= \frac{1}{secx}secxtanx                          v=e^x

du =  tanx dx

\dpi{120} \int e^{x}\left ( tanx +lnsecx \right )dx= e^{x}ln secx -\int e^{x}tanx dx +\int e^{x} tanxdx

\dpi{120} \int e^{x}\left ( tanx +lnsecx \right )dx= e^{x}ln secx +c

 

 

 

 

 

 

Practice problems:

Solve the following integrals Integration by parts(IBP) formula,

  1. \dpi{120} \int e^{x}x^{3}dx
  2.  \dpi{120} \int e^{-x}cosx dx
  3. \dpi{120} \int sin^{-1}x dx
  4. \dpi{120} \int \frac{lnx}{x^{2}}dx
  5. \dpi{120} \int e^{x}\left ( sinx+cosx \right )dx

 

 

 

 

 

 

Answers:

  1. \dpi{120} e^{x}(x^{3}-3x^{2}+6x-6)+c
  2. \dpi{120} \frac{e^{-x}}{2}(sinx-cosx)+c
  3. \dpi{120} xsin^{-1}x+\sqrt{1-x^{2}}+c
  4. \dpi{120} \frac{-1}{x}(1+lnx)+c
  5. \dpi{120} e^{x}sinx +c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Integral by Substitution

/by

Integral using Substitution

Some integrals can be reduced to standard form by proper substitution. If Ø(x)  is a continuously differentiable function, then to evaluate integrals of the form

\dpi{120} \mathbf{\int f(\varnothing (x))\varnothing '(x)dx}

We substitute Ø(x) as u  and Ø'(x)dx as du .

These  substitutions  reduce the above integral  to ∫f(u)du . After evaluating the integral  we substitute back  value of u.

 

1) Integral of the form \dpi{120} {\color{Red} \mathbf{\int f(ax+b)dx}}

\dpi{150} \mathbf{\int f(ax+b)dx=\frac{\varnothing (ax+b)}{a}+c}    where Ø is the integral of f.

\dpi{150} \mathbf{\int \left ( ax+b \right )^{n}dx=\frac{(ax+b)^{n}}{a(n+1)}+c}

 

Example1: Evaluate  \dpi{120} {\color{Red} \int (2x-3)^{5}dx}

Solution: Using the formula above we get,

\dpi{120} \int (2x-3)^{5}dx=\frac{(2x-3)^{5+1}}{2(5+1)}+c

\dpi{120} =\frac{(2x-3)^{6}}{12}+c

 

Example2: Evaluate   \dpi{120} {\color{Red} \int \frac{1}{(7x-5)^{3}}+\sqrt{3x+2}dx}

Solution: we can move  the exponential expression to the top  and rewrite radical expression with exponents  ½  to be able to apply  first formula.

\dpi{120} \int (7x-5)^{-3}+(3x+2)^{\frac{1}{2}}dx

\dpi{120} \frac{(7x-5)^{-3+1}}{7(-3+1)}+\frac{(3x+2)^{\frac{1}{2}+1}}{3(\frac{1}{2}+1)}+c

\dpi{120} \frac{(7x-5)^{-2}}{-14}+\frac{(3X+2)^{\frac{3}{2}}}{\frac{9}{2}}+C

 

2) Integral of the form  \dpi{120} \mathbf{{\color{Red} \int \frac{f'(x)}{f(x)}dx}}

If numerator  of integrand is  derivative of denominator  then its integral is log of the denominator.

\dpi{150} \mathbf{\int \frac{f'(x)}{f(x)}dx=ln|f(x)|+c}

 

Example3 : Evaluate  \dpi{120} {\color{Red} \int \frac{2x+5}{x^{2}+5x-7}dx}

Solution:               let     \dpi{120} x^{2}+5x-7 = u

2x+5  dx = du

Using this substitution we get new integral  as

\dpi{120} \int \frac{2x+5}{x^{2}+5x-7}dx =\int \frac{1}{u}du

=  ln|u| +c

= ln |x^2+5x-7|  +c

When we are solving long complex  integral problems then we can avoid all this work  directly using formula like this..

\dpi{120} \int \frac{2x+5}{x^{2}+5x-7}dx = ln|x^{2}+5x-7| +c

 

Example4: Evaluate    \dpi{120} {\color{Red} \int e^{3lnx}(x^{4}+1)^{-1}dx}

Solution: Before solving  this integral we should simplify it.

\dpi{120} \int e^{3lnx}(x^{4}+1)^{-1}dx = \int e^{lnx^{3}}\frac{1}{(x^{4}+1)}dx                                   [e and ln are inverse of each other]

\dpi{120} = \int x^{3}\frac{1}{(x^{4}+1)}dx =\int \frac{x^{3}}{x^{4}+1}dx

Now this integral is easy to solve using substitution.

Let  x^4 +1 = u

4x^3  dx =  du

\dpi{120} x^{3}dx = \frac{du}{4}

Using this substitution, new integral becomes

\dpi{120} \int \frac{x^{3}}{x^{4}+1}dx=\int \frac{1}{u}\frac{du}{4}

\dpi{120} =\frac{1}{4}\int \frac{1}{u}du

\dpi{120} =\frac{1}{4}ln|u|+c

Plug in back value of u,

\dpi{120} =\frac{1}{4}ln|x^{4}+1|+c    [Answer]

WE can  cut short this above boxed part using formula directly. After simplification we get the integral as, \dpi{120} \int \frac{x^{3}}{x^{4}+1}dx   where we can see that numerator is derivative  of denominator if 4 is added to x^3. To get 4 with x^3 we  multiply the expression with 4/4.

\dpi{120} = \frac{1}{{\color{Red} 4}}\int \frac{{\color{Red} 4}x^{3}}{x^{4}+1}dx

\dpi{120} =\frac{1}{4}ln|x^{4}+1| +c

 

3) Integral of the form \dpi{120} \mathbf{{\color{Red} \int f(x)^{n}f'(x)dx}}

For this type of integrals we use following  formula,

\dpi{150} \mathbf{\int f(x)^{n}f'(x)dx = \frac{f(x)^{n+1}}{n+1}} , n\neq -1

 

 

Example5: \dpi{120} {\color{Red} \int \frac{3x+1}{\left (3x^{2}+2x+1 \right )^{3}}dx}        

Solution:           let   3x^2+2x+1 =u

6x+2  dx =  du

2(3x+1) dx = du

(3x+1)dx = du/2

So the given integral changes to,

\dpi{120} \int \frac{3x+1}{\left (3x^{2}+2x+1 \right )^{3}}dx =\int \frac{1}{u^{3}}\frac{du}{2}

\dpi{120} \frac{1}{2}\int u^{-3} du

\dpi{120} \frac{1}{2}*\frac{u^{-2}}{-2}+c

\dpi{120} \frac{1}{-4 u^{2}}+c

\dpi{120} \frac{-1}{4(3x^{2}+2x+1)^{2}}+c

We can cut short all this work by applying formula directly, if we rewrite the given integral as

\dpi{120} \int (3x+1)(3x^{2}+2x+1)^{-3}dx

We know that derivative of    \dpi{120} 3x^{2}+2x+1    is 6x+2 so we need to multiply  this expression with 2. For that we  multiply  the expression with 2/2 and then apply the formula 3.

\dpi{120} \frac{1}{2}\int 2(3x+1)(3x^{2}+2x+1)^{-3}dx

\dpi{120} \frac{1}{2}\frac{(3x^{2}+2x+1)^{-3+1}}{-3+1}+c

\dpi{120} \frac{1}{2}\frac{(3x^{2}+2x+1)^{-2}}{-2}+c

\dpi{120} \frac{-1}{4(3x^{2}+2x+1)^{2}}+c

 

 

Example6: \dpi{120} {\color{Red} \int \frac{\sqrt{2+lnx}}{x}dx}   

Solution:  Rewriting this integral as..

\dpi{120} \int \frac{\sqrt{2+lnx}}{x}dx =\int \frac{1}{x}(2+lnx)^{\frac{1}{2}}dx

Here function f(x)=2+lnx and derivative f’(x)= 1/x given along with , so applying  formula 3 directly,

\dpi{120} \frac{(2+lnx)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c

\dpi{120} \frac{2}{3}(2+lnx)^{\frac{3}{2}}+c

 

4) Integral of the form \dpi{120} {\color{Red} \mathbf{\int sin^{m}x cos^{n}x dx}}

For this type  we follow this algorithm.

If exponent of sinx (m) is odd then put  cosx=u

If exponent of cosx (n) is odd then put sinx=u

If exponents of sinx and cosx both are  odd then put either sinx=u or cosx=u.

If exponents of sinx and cosx  both are even then express  \dpi{120} sin^{m}xcos^{n}x   in terms of sines and cosines of multiples of x by using trigonometric  results or De’ Moivere’s  theorem.

 

Example7:  \dpi{120} {\color{Red} \int sin^{3}xcos^{4}x dx}

Solution: Exponent of  sinx is odd. So we substitute

cos x= u

-sinx dx= du

sinx dx =-du

\dpi{120} \int sin^{3}xcos^{4}x dx=\int sin^{2}xcos^{4}x sinx dx

\dpi{120} \int (1-cos^{2}x)cos^{4}x sinx dx

\dpi{120} \int (1-u^{2})u^{4} (-du)

\dpi{120} -\int (u^{4}-u^{6})du

\dpi{120} -\frac{u^{5}}{5} +\frac{u^{6}}{6}+c

 

\dpi{120} -\frac{cos^{5}x}{5} +\frac{cos^{6}x}{6}+c

 

 

 

Example8:  \dpi{120} {\color{Red} \int sin^{2}x cos^{5}x dx}       

Solution: Exponents of cosx is odd so we substitute sinx=u

sinx  =  u

cosx dx = du

\dpi{120} \int sin^{2}x cos^{5}x dx = \int sin^{2}x cos^{4}x cosx dx

 

\dpi{120} \int sin^{2}x(1-sin^{2}x)^{2}cosx dx

\dpi{120} \int u^{2}(1-u^{2})^{2}du

\dpi{120} \int u^{2}(1-2u^{2}+u^{4})du

\dpi{120} \int u^{2}-2u^{4}+u^{6}du

\dpi{120} \frac{u^{3}}{3}-\frac{2}{5}u^{5}+\frac{u^{7}}{7}+c

Finally plugin back u as sinx and get  answer.

\dpi{120} \frac{sin^{3}x}{3}-\frac{2}{5}sin^{5}x+\frac{sin^{7}x}{7}+c

 

 

 

 

 

Practice problems: Solve the following integrals using substitution.

1.\dpi{120} \int \frac{e^{3x}}{e^{3x}+1}dx

2. \dpi{120} \int \frac{1}{x(3+lnx)}dx

3. \dpi{120} \int x\sqrt{x+2}dx

4. \dpi{120} \int \frac{(lnx)^{3}}{x}dx

5. \dpi{120} \int \frac{(x+1)e^{x}}{cos^{2}(xe^{x})}dx

 

 

 

 

 

Answers:

  1. \dpi{120} \frac{1}{3}ln|e^{3x}+1|+c
  2. \dpi{120} ln|3+lnx|+c
  3.  \dpi{120} \frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+c
  4. \dpi{120} \frac{(lnx)^{4}}{4}+c
  5. \dpi{120} tan(xe^{x})+c

Indefinite Integral Using Fundamental Rules

/by

Fundamental rules of Integration

We need to be familiar with all the following rules  and formulas to learn basic integration techniques.

  1. \dpi{120} \mathbf{\int x^{n}dx = \frac{x^{n+1}}{n+1}+c ,n\neq -1}          [called power rule of integration]
  2. \dpi{120} \mathbf{\int \frac{1}{x}dx= ln|x| +c}
  3.  \dpi{120} \mathbf{\int e^{x}dx = e^{x}+c}
  4. \dpi{120} \mathbf{\int a^{x}dx=\frac{a^{x}}{lna}+c}
  5. \dpi{120} \mathbf{\int sin(x) dx = -cos(x)+c}
  6. \dpi{120} \mathbf{\int cos(x)dx = sin(x)+c}
  7.  \dpi{120} \mathbf{\int sec^{2}(x)dx = tan(x)+c}
  8. \dpi{120} \mathbf{\int Csc^{2}(x)dx =-cot(x)+c}
  9. \dpi{120} \mathbf{\int sec(x)tan(x)dx =sec(x)+c}
  10. \dpi{120} \mathbf{\int csc(x)cot(x)dx =-csc(x)+c}
  11. \dpi{120} \mathbf{\int cot(x)dx = ln|sinx|+c}
  12. \dpi{120} \mathbf{\int tan(x)dx = -ln|cosx| +c}
  13. \dpi{120} \mathbf{\int sec(x)dx = ln|secx +tanx| +c}
  14. \dpi{120} \mathbf{\int csc(x)dx = ln|cscx-cotx| +c}

.

 

Sum  or Difference of  different functions

\dpi{150} \mathbf{\int \left [f(x)\pm g(x) \right ] dx = \int f(x)dx\pm \int g(x)dx}

Integral of the sum or difference of a finite number of functions is equal to sum or difference of integrals of various functions.

In simple words we can take integrals of each function separately.

Lets work on some examples using these rules.

 

Example1:  Evaluate

\dpi{120} {\color{Red} \int \left (x^{4}+\sqrt{x} \right )dx}

Solution:

\dpi{120} \int \left (x^{4}+\sqrt{x} \right )dx = \int x^{4}dx+\int x^{\frac{1}{2}}dx

\dpi{120} =\frac{x^{4+1}}{4+1}+\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+ c

\dpi{120} =\frac{x^{5}}{5}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}} +c

\dpi{120} =\frac{x^{5}}{5}+\frac{2x^{\frac{3}{2}}}{3} +c

 

Example2: Evaluate the following indefinite integral.

\dpi{120} {\color{Red} \int \frac{1}{\sqrt{x}}+\frac{1}{x^{3}}dx}

Solution:  Take all terms to the top and rewrite using negative exponents so that we can apply power rule of integration.

\dpi{120} \int \frac{1}{\sqrt{x}}+\frac{1}{x^{3}}dx = \int x^{\frac{-1}{2}}+x^{-3}dx

\dpi{120} =\frac{x^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+ \frac{x^{-3+1}}{-3+1}+c

\dpi{120} =\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\frac{x^{-2}}{-2}+c

\dpi{120} =2\sqrt{x}-\frac{1}{2x^{2}}+c                                          [answer]

 

Example3: \dpi{120} {\color{Red} \int \frac{2}{x}+\frac{x}{2}+x^{2}+2^{x} dx}

Solution: Rewrite the given expression after taking out constants.

\dpi{120} \int \frac{2}{x}+\frac{x}{2}+x^{2}+2^{x} dx = 2\int \frac{1}{x}dx+\frac{1}{2}\int x dx +\int x^{2}dx +\int 2^{x}dx

\dpi{120} = 2ln|x| +\frac{1}{2}*\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{2^{x}}{ln2}+c

\dpi{120} = 2ln|x| +\frac{x^{2}}{4}+\frac{x^{3}}{3}+\frac{2^{x}}{ln2}+c

 

Example4:    \dpi{120} {\color{Red} \int e^{x}+tan^{2}(x) dx}

Solution:     \dpi{120} \int e^{x}+tan^{2}(x) dx=\int e^{x}+\left (sec^{2}x-1 \right ) dx

\dpi{120} =\int e^{x}dx +\int sec^{2}x dx - \int 1 dx

\dpi{120} = e^{x}+tanx -x+c

 

Example5:   \dpi{120} {\color{Red} \int \frac{x\left (2^{x} \right )+1}{x}dx}

Solution:

\dpi{120} \int \frac{x\left (2^{x} \right )+1}{x}dx = \int \frac{x(2^{x})}{x}+\frac{1}{x} dx

\dpi{120} =\int 2^{x}dx + \int \frac{1}{x}dx

\dpi{120} =\frac{2^{x}}{ln2}+ln|x| +c

 

Example6:  \dpi{120} {\color{Red} \int \frac{sinx}{1+sinx} dx}

Solution:

\dpi{120} \int \frac{sinx}{1+sinx} dx = \int \frac{sinx}{1+sinx}*\frac{1-sinx}{1-sinx}dx

\dpi{120} =\int \frac{sinx(1-sinx)}{(1+sinx)(1-sinx)}dx

\dpi{120} =\int \frac{sinx-sin^{2}x}{1-sin^{2}x}dx

\dpi{120} =\int \frac{sinx-sin^{2}x}{cos^{2}x}dx

\dpi{120} =\int \frac{sinx}{cos^{2}x}-\frac{sin^{2}x}{cos^{2}x}dx

\dpi{120} =\int secxtanx - tan^{2}x dx

\dpi{120} =\int secxtanx - \left (sec^{2}x-1 \right ) dx

\dpi{120} =\int secxtanx dx- \int sec^{2}xdx +\int 1 dx

= sec x – tanx + x+ c

 

 

 

Practice problems:

Solve the following indefinite integrals using  fundamental rules.

  1. \dpi{120} \int \frac{x^{2}-3x+2}{x}dx
  2. ⎰(1-x)√x dx
  3. \dpi{120} \int \frac{1}{\sqrt[3]{x^{2}}}dx
  4. \dpi{120} \int 2^{x}+\frac{5}{x}-\frac{1}{x^{\frac{1}{3}}}dx
  5. \dpi{120} \int \frac{1}{1+cosx}dx

 

Answers:

  • (x^2)/2 -3x +2ln|x|+c
  • \dpi{120} \frac{2}{3}x^{\frac{3}{2}}-\frac{2}{5}x^{\frac{5}{2}}+c
  • 3x^(1/3)+c
  • \dpi{120} \frac{2^{x}}{ln2}+5ln|x|-\frac{3}{2}x^{\frac{2}{3}}+c
  • -cotx +cscx +c

 

 

Definite Integrals and its properties

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How to solve definite Integral 

To solve definite integral we make use of fundamental theorem of Calculus which says that if a function f is continuous on close interval [a,b] and F is the indefinite integral of f on this interval then,

\dpi{120} \mathbf{\int_{a}^{b}f(x)dx = F(b)-F(a)}

This is also called first fundamental theorem of calculus.

While calculating definite integrals, we don’t keep constant of integration ‘C’

In words we can explain definite integral as,

\dpi{120} \int_{a}^{b}f(x)dx =  Value of integral at upper limit (b) – value of integral at lower limit (a)

Lets understand this process using a simple example.

 

Example1: Evaluate \dpi{120} {\color{Red} \int_{1}^{2}x^{2} dx}

Solution: