Antiderivative.mws

Indefinite Integrals

>    restart;
with( plots ): 

Warning, the name changecoords has been redefined

>   

Lesson Overview

All of our attention to date has been on the computation of derivatives.  This unit turns things unpside down.  Now, instead of finding the derivative a function you will be asked to identify a function based on information about its derivative.  From the perspective of differentiation these functions are known as antiderivatives .  (However, because the stem anti  means not  and the stem retro  means undo , it seems that a more appropriate term would be retroderivative.)  From the perspective of area (to be introduced later in the Riemann Sum and Definite Integral lessons of this unit) it makes more sense to talk about the indefinite integral  of a function.

The goals of this lesson are

The graphical understanding of antiderivatives is greatly enhanced with the use of the Antiderivative  maplet [ Maplet Viewer][ MapleNet].

>   

Definition (Antiderivative/Indefinite Integral)

A function F is an antiderivative , or indefinite integral , of f on an interval I  if

   diff(F(x),x) = f(x)   for all x  in I .  

Notes

  • The definition provides a direct check to determine if F is, in fact, an antiderivative of f:

  differentiate F (with respect to x ) and compare this result with f on the interval I   

  • A common alternate notation for antiderivatives is to use integral notation  

   Diff(F(x),x) = f(x)   if and only if   F(x) = Int(f(x),x) .  

  • In the expression Int(f(x),x) , read "the indefinite integral of f(x)  with respect to x ", the first symbol (an elongated S) is called the integral sign , the integrand  is f(x) , and x  is the variable of integration .  The integral sign and the dx should be viewed as a matched pair -- much like left- and right-parentheses.
  • The definition and these notes are carefully written to state that F is an antiderivative of f (and not F is the antiderivative of f).  The reason for this will be made very clear in the examples.

>   

Example 1:   Int(3*x^2-2,x) = x^3-2*x+C   

Let

>    f1 := 3*x^2 - 2:
f(x) = f1;

f(x) = 3*x^2-2

Consider the functions

>    F1 := x^3 - 2*x:
F[1](x) = F1;

F[1](x) = x^3-2*x

>    F2 := x^3 - 3*x + 2:
F[2](x) = F2;

F[2](x) = x^3-3*x+2

>    F3:= x^3 - 2*x + 3:
F[3](x) = F3;

F[3](x) = x^3-2*x+3

Analytic Verification

From the derivatives of these functions:

>    dF1 := Diff( F1, x ):
dF1 = value( dF1 );

Diff(x^3-2*x,x) = 3*x^2-2

>    dF2 := Diff( F2, x ):
dF2 = value( dF2 );

Diff(x^3-3*x+2,x) = 3*x^2-3

>    dF3 := Diff( F3, x ):
dF3 = value( dF3 );

Diff(x^3-2*x+3,x) = 3*x^2-2

it is seen that F ``[1]  and F ``[3]  are antiderivatives (indefinite integrals) of f and that F ``[2]  is not  an antiderivative of f.

>   

Graphical Verification

To understand why there can be more than one antiderivative of f(x) = 3*x^2-2 , consider the graph of F ``[1] , F ``[2] , and F ``[3]  on the interval [ -2, 2 ]:

>    plot( [ F1, F2, F3 ], x=-2..2,
      color=[red,blue,green], legend=["y=F1", "y=F2", "y=F3" ] );

[Maple Plot]

Observe that the graph of y  = F ``[3]  is the graph of y  = F ``[1]  shifted vertically by 3, i.e., F ``[3] ( x ) = F ``[1] ( x ) + 3.  The graph of y  = F ``[2]  is not a vertical shift of the other graphs.  In general, any vertical shift of an antiderivative is sitll an antiderivative.  This is a direct consequence of the fact that the derivative of a constant is zero.  In the case of the specific functions in this example, any function of the form F(x) = x^3-2*x+C , where C  is a constant, is an antiderivative of f(x) = 3*x^2-2 .  In the notation for indefinite integrals,

   Int(3*x^2-2,x) = x^3-2*x+C   

>   

The Antiderivative  Maplet

Further verification of this fact can be obtained with the Antiderivative  maplet.  If Maple is installed on your local computer, launch the maplet with the Maplet Viewer; otherwise, use MapleNet.  Change the Function  to 3*x^2-2 ; leave a  and b  unchanged.  Click the Plot  button.  The Plot Window will display a graph of the function y = 3*x^2-2  and one antiderivative y = x^3-2*x+4  (the antiderivative that is zero at x = a ).  Now, click the check box for Show class of antiderivatives  and then click the Plot  button.  The Plot Window now displays 20 antiderivatives of y=3*x^2-2.  Notice how all 20 green curves are vertical shifts of the blue curve.

>   

>   

The Antiderivative Rules

Every Derivative Rule has a corresponding Antiderivative Rule.

   MATRIX([[Name, `Derivative Rule`, `Antiderivative Rule`, `Extra Conditions`], [___________________, __________________________, _________________________________, ______________], [Constant, Diff(k,x) ...

  

The restriction that r <> -1  for the Power and Generalized Power Rules is necessary because we do not (yet) know any function whose derivative is x^(-1) = 1/x .  The most likely candidate is x^0 = 1 , but we know diff(x^0,x) = 0 .  If you ever forget this restriction and attempt to apply the Power or Generalized Power Rule with r = -1 , division by r+1 = 0  will remind you that this is not possible.

The antiderivative rules obtained from the quotient, product, and chain rules are not of interest at this time.  The remaining examples in this lesson demonstrate the application of these rules.

>   

Example 2: Int(3*x^2-2,x)  

This is the same indefinite integral considered in Example 1.  This time, let's determine the family of all antiderivatives using the Antiderivative Rules.

   MATRIX([[Int(3*x^2-2,x), `Original Problem`], [`` = Int(3*x^2,x)-Int(2,x), Difference*Rule], [`` = 3*Int(x^2,x)-2*Int(1,x), `Constant Multiple Rule (twice)`], [`             ` = 3*(x^3/3+C[1])-2*(x+C[2...   

Note that the constants could have been omitted when the Power and Identity Rules were applied provided the constant is included at the end.

>   

Example 3: Int(2/(x^4)+sqrt(x)+3*sin(x),x)  

The Antiderivative Rules can be applied to evaluate this indefinite integral as follows:

   MATRIX([[Int(2/(x^4)+sqrt(x)+3*sin(x),x), `Original Problem`], [`      ` = 2*Int(x^(-4),x)+Int(x^(1/2),x)+3*Int(sin(x),x), `Sum and Constant Multiple Rules`], [`` = 2*x^(-3)/(-3)+x^(3/2)/(3/2)+3*(-cos(...   

To confirm this result simply compute the derivative of this answer:

>    ans3 := -2/(3*x^3) + 2/3*x^(3/2) - 3*cos(x) + C;

ans3 := -2/3/x^3+2/3*x^(3/2)-3*cos(x)+C

>    dF3 := Diff( ans3, x ):
dF3 = value( dF3 );

Diff(-2/3/x^3+2/3*x^(3/2)-3*cos(x)+C,x) = 2/x^4+x^(1/2)+3*sin(x)

Because the derivative of our answer is the original function, the antiderivative is correct!  That is,

   Int(2/(x^4)+sqrt(x)+3*sin(x),x) = -2/(3*x^3)+2/3*x^(3/2)-3*cos(x)+C .  

Note that if the constant C  is omitted, then we would have an  antiderivative.  The same would be true if a specific value were assigned to C .  When we say the  antiderivative, we mean the family of all antiderivatives.  Only when the constant is allowed to take on all real values have we found the  antiderivative of the function.

>   

Example 4: Int((t-1)^2/sqrt(t),t)  

First, note that the dt indicates that the t is the independent variable in this problem.  Before the antiderivative Rules can be applied to this problem it is necessary to expand and simplify the integrand.  A full explanation of the steps to find all antiderivatives of the integrand are contained below:

   MATRIX([[Int((t-1)^2/sqrt(t),t), `Original Problem`], [`` = Int((t^2-2*t+1)/(t^(1/2)),t), `algebraic expansion`], [`                      ` = Int(t^(3/2)-2*t^(1/2)+t^(-1/2),t), `more algebra`], [`     ...   

To confirm this result, let

>    ans4 := 2/5*t^(5/2)-2*2/3*t^(3/2)+2*t^(1/2)+C;

ans4 := 2/5*t^(5/2)-4/3*t^(3/2)+2*t^(1/2)+C

Then, computing the derivative with respect to t :

>    dF4 := Diff( ans4, t ):
dF4 = value( dF4 );
`` = factor( value(dF4) );

Diff(2/5*t^(5/2)-4/3*t^(3/2)+2*t^(1/2)+C,t) = t^(3/2)-2*t^(1/2)+1/(t^(1/2))

`` = (t-1)^2/t^(1/2)

Thus,

   Int((t-1)^2/sqrt(t),t) = 2/5*t^(5/2)-4/3*t^(3/2)+2*t^(1/2)+C .  

>   

Example 5:   Int((5*x^4+2*x)*sqrt(x^5+x^2+2),x)  

There is not much that can be done to rewrite the integrand of this indefinite integral in a form so that the Sum, Difference, and Constant Multiple Antiderivative Rules can be applied.  But, notice that the coefficient of the square root is exactly the derivative of the expression inside the square root.  This is the pattern that must exist to use the Generalized Power Rule with g(x) = x^5 + x^2 + 2 and r=1/2:

   MATRIX([[Int((5*x^4+2*x)*sqrt(x^5+x^2+2),x), `Original Problem`], [`               ` = Int(diff(g(x),x)*g(x)^(1/2),x), `substitute: `*g(x) = x^5+x^2+2], [`` = g(x)^(3/2)/(3/2)+C, `Gen'l Power Rule, r=1...

Note that the constant appears in the problem when the integral sign disappears.  And that it appears outside all other expressions.

As before, this result can be confirmed by differentiation:

>    ans5 := 2/3*(x^5+x^2+2)^(3/2)+C;

ans5 := 2/3*(x^5+x^2+2)^(3/2)+C

>    dF5 := Diff( ans5, x ):
dF5 = value( dF5 );

Diff(2/3*(x^5+x^2+2)^(3/2)+C,x) = (x^5+x^2+2)^(1/2)*(5*x^4+2*x)

Great!  This shows that

   Int((5*x^4+2*x)*sqrt(x^5+x^2+2),x) = 2/3*(x^5+x^2+2)^(3/2)+C .  

>   

Example 6: Int(z/((z^2+1)^2),z)  

Compare this integrand to the integrand in Example 4.  Note that expanding the denominator of the integrand does not help to find the antiderivative.  Instead, it should be noted that if the numerator is 2*z then it would be the derivative of ( z^2+1 ).  Then, writing the 1/(z^2+1)^2 with a negative exponent, the Generalized Power Rule could be used to evaluate this indefinite integral.  Here is the complete evaluation:

   MATRIX([[Int(z/((z^2+1)^2),z), `Original Problem`], [`` = Int(z*(z^2+1)^(-2),z), `rewrite with negative exponent`], [`       ` = 1/2*Int(2*z*(z^2+1)^(-2),z), `multiply by `*1 = 1/2*2], [`            ` ...   

>   

Lesson Summary

The process of undoing differentiation is called antidifferentiation or indefinite integration.  Because the derivative of a constant is zero, diff(C,x) = 0 , if one antiderivative can be found, then adding an arbitrary constant produces an infinite family of antiderivatives.  The Antiderivative Rules are obtained by reversing a Differentiation Rule.  One nice property of antiderivatives is that it is always possible to use differentiation to check that an antiderivative is correct.

>   

What's Next?

The ability to find antiderivatives and evaluate indefinite integrals is fundamental to the final two units of this course.  The Antiderivative  maplet [ Maplet Viewer][ MapleNet] can assist with your development of a better understanding of the geometric relationship between a function and its antiderivatives.  Use the online practice sessions for this lesson to improve your skills.  The general practice session should be mastered prior to attempting the online homework assignment.  The assigned problems from the text should also be completed.

The next lesson applies the search for antiderivatives to the search for solutions to Differential Equations.  This lesson is a very brief introduction to an important topic in mathematics.  Many of you will probably take a course in Differential Equations after you complete the Calculus sequence.

>   

>