EvaluateDefInt.mws

Evaluating Definite Integrals

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with( plots ): 

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Lesson Overview

This lesson could be the first lesson in the Applications of the Definite Integral unit.  It is included here as a way to obtain additional practice with the evaluation of definite integrals.  The main new technique that will be introduced is the Method of Substitution; an integral form for the Chain Rule.

The examples provide a variety of different integration problems.  General problems involving representative integrands for definite (and indefinite) integrals, generic even and odd functions, and the derivation of an antiderivative of 1/x  and properties of this antiderivative are discussed in detail.

Remember to use the Integration  maplet [ Maplet Viewer][ Maplenet] to help develop your integration skills.

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Method of Substitution

Recall the Chain Rule (from the Differentiation Rules lesson):  If y = f(u)  and u = g(x) , where f and g are both differentiable functions, then

   dy/dx = Diff(f(g(x)),x)  = `f '`(g(x))*`g'`(x) .  

The integral form of this result is:

   Int(`f '`(g(x))*`g'`(x),x)  = f(g(x))+C .  

Of course, for a definite integral the Second Fundamental Theorem of Calculus provides:

   Int(`f '`(g(x))*`g'`(x),x = a .. b)  = f(g(b))-f(g(a)) .  

The key is to identify the integrand as the product of an expression that involves g(x)  and the derivative `g'`(x) .  In practice, one typically rewrites the integrand using the substitution u = g(x)  and replacing the differential `g'`(x)*dx  with du :

   Int(`f '`(g(x))*`g'`(x),x) = Int(`f '`(u),u) .  

This puts the integral in a form where it is easy to identify an antiderivative:

   Int(`f '`(u),u)  = f(u)+C .  

All that remains is to use the substitution to rewrite the antiderivative in terms of the original independent variable:

   f(u)+C  = f(g(x))+C .  

When the same steps are repeated for a definite integral the method can be written as

   Int(`f '`(g(x))*`g'`(x),x = a .. b) = Int(`f '`(u),u = g(a) .. g(b))  = f(g(b))-f(b(a)) .  

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Example 1: Indefinite Integral

Evaluate the indefinite integral

   Int(2*x/sqrt(1+x^2),x) .  

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In the integrand notice that the numerator 2*x  is the derivative of x^2 .  While x^2  does appear elsewhere in the intgrand this observation is not immediately useful as an antiderivative of 1/sqrt(1+u)  is not immediately known.  But, 2*x  is also the derivative of 1+x^2  and an antiderivative of 1/sqrt(u) = u^(-1/2)  is known   Int(u^(-1/2),u) = 2*u^(1/2) .

To put this together to complete the evaluation of this indefinite integral with the Method of Substitution:

     Let u = 1+x^2 .

     Then du = 2*x*dx  and

   Int(2*x/sqrt(1+x^2),x) = Int(1/sqrt(u),u)                    

               = Int(u^(-1/2),u)   

             = 2*u^(1/2)+C   

                    = 2*sqrt(1+x^2)+C .  

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As with any indefinite integral, this answer can be checked:

   Diff(2*sqrt(1+x^2)+C,x) = 2/2*(1+x^2)^(-1/2)*2*x+0      

         = 2*x/sqrt(1+x^2) .  

The fact that the derivative of the general antiderivative is the original integrand means that the computation has been done correctly.

Solution using the Integration  maplet [ Maplet Viewer][ Maplenet]

The Integration  maplet [ Maplet Viewer][ Maplenet] can be used to help find the antiderivative in this problem.  Once the maplet is launched enter 2*x/sqrt(1+x^2)  in the Function  field and x  in the Variable  field (as this is an indefinite integral, leave the from  and to  fields empty).

Click the Start  button.  To see the effect of trying the substitution u = x^2 , enter u=x^2 , u in the large box to the right of the Rewrite  and Change  buttons, then click Change .  The Problem Status  window will show that this substitution converts the integral to Int((1+u)^(1/2),u) .  As this is not a form for any of our known antiderivatives, this is a deadend for us.

To restart the problem, click the Start  button.  To see the effect of trying the substitution u = 1+x^2 , enter u=1+x^2, u  in the large box to the right of the Rewrite  and Change  buttons, then click Change .  The Problem Status  window will show that this substitution converts the integral to Int(u^(1/2),u)  --- success!  Click the Power  button to have the Power Rule applied.  After this step the indefinite integral is evaluated as sqrt(u) .  To complete the evaluation of the indefinite integral it is necessary to undo the substitution; to do this click the Revert  button.  When writing the answer on paper do not forget to add the constant of integration.

Note

If you ask the Integration  maplet to show All Steps in the evaluation of this integral the results will be different from --- but ultimately equivalent to --- what has been discussed in this lesson.

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Example 2: Definite Integral

Evaluate the definite integral

   Int(cos(3*theta),theta = -Pi/6 .. Pi/3) .  

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The only form of the cosine function that has a known antiderivative is Int(cos(u),u) = sin(u)+C .  In order to put the given integral in a form where this can be used, let u = 3*theta .  Then du = 3*d*theta .  While there is no factor in the integrand that matches the coefficient of 3 in the equation for the differentials, it is possible to replace d*theta  with du/3  as follows:

   Int(cos(3*theta),theta = -Pi/6 .. Pi/3) = Int(cos(u)*`(1/3)`,u = -Pi/2 .. Pi)           

                 = 1/3   Int(cos(u),u = -Pi/2 .. Pi)   

                          = 1/3  ( sin(Pi)-sin(-Pi/2)  )  

                 = 1/3  ( 0-(-sqrt(2)/2)  )  

     = sqrt(2)/6 .           

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Notice that implicit in this work is the result of the indefinite integral:

   Int(cos(3*theta),theta) = Int(cos(u)*`(1/3)`,u)   

     = 1/3  Int( cos(u), u )  

     = 1/3   sin(u)  + C   

     = 1/3   sin(3*theta)  + C  

and the definite integral can be evaluated using the Second Fundamental Theorem of Calculus:

                 Int(cos(3*theta),theta = -Pi/6 .. Pi/3)  = 1/3   sin(3*Pi/3)  - 1/3   sin(3*(-Pi/6))  )  

                         = 1/3   sin(Pi)  - 1/3   sin(-Pi/2)   

           = 0 - 1/3  ( -sqrt(2)/2 )  

     = -sqrt(2)/6        

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Example 3: Odd Function on a Symmetric Interval

Recall that an odd function is a function with the property that f(-x) = -f(x) .  An odd function is said to be symmetric about the origin.  When an odd function is integrated over a symmetric interval [ -a , a  ] ( a  > 0 ) the value of the integral can be simplified.  Suppose that f(x)  > 0 for x  > 0.  Then, by symmetry, the "area" under the curve on the interval [ -a , 0 ] is the negative of the area under the curve on the interval [ 0, a  ].

>    P1 := plot( x+sin(3*x), x=-Pi..0, filled=true, color=cyan ):
P2 := plot( x+sin(3*x), x= 0..Pi, filled=true, color=pink ):
display( P1,P2, title="Odd Function: Areas Cancel" );

[Maple Plot]

In the above graph notice how the positive pink area exactly cancels the negative cyan area.

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This suggests that

   Int(f(x),x = -a .. a) = 0 .  

To prove this fact for any (integrable) odd function, separate the original interval [ -a , a  ] into two region [ -a , 0 ] and [ 0, a  ].  For the integral on [ -a , 0 ] use the substitution u = -x .  Then x = -u  and dx = -du  and the resulting integral is:

   Int(f(x),x = -a .. 0) = Int(f(-u)*`(-1)`,u = a .. 0)         ( Substitution )    

          = -Int(f(-u),u = a .. 0)              ( Algebra )

                                             = -Int(-f(u),u = a .. 0)             ( Definition of Odd Function )       

          = Int(f(u),u = a .. 0)                 ( Algebra )

                                        = -Int(f(u),u = 0 .. a)              ( Property of Definite Integral )

Then, recalling that the integration variable in a definite integral does not make a difference to the value of the integral

   Int(f(x),x = -a .. a) = Int(f(x),x = -a .. 0)+Int(f(x),x = 0 .. a)   

                    = -Int(f(u),u = 0 .. a)+Int(f(x),x = 0 .. a)   

     = 0.                       

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One example of how this fact could be used to evaluate a definite integral of an odd function over a non-symmetric interval is as follows::  To evaluate Int(x^7-x^3+x-1,x = -1/2 .. 1)  one could use the fact that the integrand is an odd function to reduce the problem over a shorter interval:

   Int(x^7-x^3+x-1,x = -1/2 .. 1) = Int(x^7-x^3+x-1,x = -1/2 .. 1/2)+Int(x^7-x^3+x-1,x = 1/2 .. 1)   

     = 0+Int(x^7-x^3+x-1,x = 1/2 .. 1)   

This computation is somewhat simpler in that it does not require handling all the negative signs that could arise from a lower limit of x = -1/2 .

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Example 4: Even Function on a Symmetric Interval

Recall that an even function is a function with the property that f(-x) = f(x) .  An even function is said to be symmetric about the y -axis.  When an even function is integrated over a symmetric interval [ -a , a  ] ( a  > 0 ) the value of the integral can be simplified.  Suppose that f(x)  > 0 for x  > 0.  Then, by symmetry, the "area" under the curve on the interval [ -a , 0 ] is the same as the area under the curve on the interval [ 0, a  ].

>    P3 := plot( abs(x+sin(3*x)), x=-Pi..0, filled=true, color=cyan ):
display( P3,P2, title="Even Function: Areas Match" );

[Maple Plot]

In the above graph notice how the positive pink area exactly matches the negative cyan area.

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This suggests that

   Int(f(x),x = -a .. a) = 2*Int(f(x),x = 0 .. a) .  

To prove this fact for any (integrable) even function, separate the original interval [ -a , a  ] into two region [ -a , 0 ] and [ 0, a  ].  For the integral on [ -a , 0 ] use the substitution u = -x .  Then x = -u  and dx = -du  and the resulting integral is:

   Int(f(x),x = -a .. 0) = Int(f(-u)*`(-1)`,u = a .. 0)         ( Substitution )    

          = -Int(f(-u),u = a .. 0)              ( Algebra )

                                              = -Int(f(u),u = a .. 0)                ( Definition of Even Function )        

          = -Int(f(u),u = a .. 0)                ( Algebra )

                                         = Int(f(u),u = 0 .. a)                  ( Property of Definite Integral )

Then, recalling that the integration variable in a definite integral does not make a difference to the value of the integral

   Int(f(x),x = -a .. a) = Int(f(x),x = -a .. 0)+Int(f(x),x = 0 .. a)   

                    = Int(f(u),u = 0 .. a)+Int(f(x),x = 0 .. a)   

  = 2*Int(f(x),x = 0 .. a)   

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Example 5: Antiderivative for 1/x  

Recall that there is one power for which we do not know an antiderivative:

   Int(x^n,x) = x^(n+1)/(n+1)+C ,   for all n <> -1 .  

Now, because x^(-1) = 1/x  is continuous for all x  > 0, we know this function is Riemann integrable on ( 0, infinity  ).  The First Fundamental Theorem of Calculus ensures that the antiderivative exists and can be written in terms of a definite integral:

>    f5 := 1/t:
F5 := Int( f5, t=1..x ):
L(x) = F5;

L(x) = Int(1/t,t = 1 .. x)

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Let's discover a few properties of this antiderivative.

Property 1: L(1) = 0  

The properties for Definite Integrals immediately tell us one function value for this function:

   L(1) = Int(1/t,t = 1 .. 1)  = 0.  

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Property 2: L(1/x) = -L(x)  

This property requires the use of the Method of Substitution with u=1/t.  Then t=1/u, dt = -u^(-2)*du, and the limits of integration are converted from t=1 to u=1 and t=1/x to u=x.  Thus

   L(1/x) = Int(1/t,t = 1 .. 1/x)       

                       = Int(u*(-u^(-2)),u = 1 .. x)   

         = -Int(1/u,u = 1 .. x)   

     = -L(x) .  

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Property 3: L(xy) = L(x)+L(y)  

Let x  and y  be positive numbers.  Then, with a conveniently selected splitting of the interval [ 1, x*y  ] into [ 1, x  ] `union`(``,``) [ x , x*y  ].

  : L(x*y) = Int(1/t,t = 1 .. x*y)                  

               = Int(1/t,t = 1 .. x)+Int(1/t,t = x .. x*y)   

          = L(x)+Int(1/t,t = x .. x*y)   

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The remaining unidentified definite integral can be simplified with the substitution u = t/x .  Then t = x*u , dt = x*du  and the limits of integration are converted from t = x  to u = 1  and t = x*y  to u = y .  When all of these facts are collected we find

   Int(1/t,t = x .. x*y) = Int(x/(x*u),u = 1 .. y)         

      = Int(1/u,u = 1 .. y)   

     = L(y) .  

Thus, for all positive values of x  and y :

   L(x*y) = L(x)+L(y) .  

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Property 4: L(x/y) = L(x)-L(y)  

Properties 3 and 2 combine to yield this result:

   L(x/y) = L(x*1/y)  = L(x)+L(1/y) = L(x)-L(y) .  

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These properties are reminiscent of the properties of the (natural) logarithm function.  To further justify this association, observe that the function L is increasing for all x  > 0 because `L'`(x) = 1/x  > 0.  In particular, L(x) > 0 for all x  > 1 and L(x)  < 0 for all 0 < x  < 1.  While L has no critical points, the fact that `L''`(x) = -1/x^2 < 0, the graph is always concave down.  All of these features are present in the following graph of L

>    plot( F5, x=0..10, title="L(x) = ln(x)" );

[Maple Plot]

All of the properties and the above graph lead us to conclude that L(x) = ln(x) , the natural logarithm.

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Lesson Summary

The Method of Substitution is the one integration technique considered in this course.  This method provides an antiderivative version of the Power Rule for differentiation.  This method greatly expands the integrals that can be compute explicitly.  Additional methods, particularly a antiderivative analog of the Product Rule, will be developed in the next course in the calculus sequence.

While the natural logarithm is not generally discussed in this course, the Method of Substitution and ideas from the Graphing lesson strongly suggest that the natural logarithm is an antiderivative for 1/x :

   ln(x) = Int(1/t,t = 1 .. x) .  

What's Next?

Now that you have completed this lesson, you should hone your integration (and antidifferentiation) skills with the online homework assignment.  The Integration maplet [ Maplet Viewer][ Maplenet] can be very helpful when exploring how different substitutions change an integral.  There are practice sessions for specific types of integrals, a general practice session for problems related to this lesson, and a graded homework assignment.  Work enough practice problems to master these concepts, then complete the online homework assignment and the assigned problems from the text.

The Definite Integrals, Fundamental Theorems of Calculus, and Evaluating Definite Integrals lessons are the source of questions for Quiz 8.

The last unit in this course presents a collection of Applications of the Integral.  The applications begin with the theoretical Mean Value Theorem for Integrals, then consider three topics from geometry --- Area of a Plane Region, Volume of a Solid, Length of a Plane Curve, before concluding with one application from physics --- Work.

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