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Calculus Volume 2

3.1 Integration by Parts

Calculus Volume 23.1 Integration by Parts
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 3.1.1. Recognize when to use integration by parts.
  • 3.1.2. Use the integration-by-parts formula to solve integration problems.
  • 3.1.3. Use the integration-by-parts formula for definite integrals.

By now we have a fairly thorough procedure for how to evaluate many basic integrals. However, although we can integrate xsin(x2)dxxsin(x2)dx by using the substitution, u=x2,u=x2, something as simple looking as xsinxdxxsinxdx defies us. Many students want to know whether there is a product rule for integration. There isn’t, but there is a technique based on the product rule for differentiation that allows us to exchange one integral for another. We call this technique integration by parts.

The Integration-by-Parts Formula

If, h(x)=f(x)g(x),h(x)=f(x)g(x), then by using the product rule, we obtain h(x)=f(x)g(x)+g(x)f(x).h(x)=f(x)g(x)+g(x)f(x). Although at first it may seem counterproductive, let’s now integrate both sides of this equation: h(x)dx=(g(x)f(x)+f(x)g(x))dx.h(x)dx=(g(x)f(x)+f(x)g(x))dx.

This gives us

h(x)=f(x)g(x)=g(x)f(x)dx+f(x)g(x)dx.h(x)=f(x)g(x)=g(x)f(x)dx+f(x)g(x)dx.

Now we solve for f(x)g(x)dx:f(x)g(x)dx:

f(x)g(x)dx=f(x)g(x)g(x)f(x)dx.f(x)g(x)dx=f(x)g(x)g(x)f(x)dx.

By making the substitutions u=f(x)u=f(x) and v=g(x),v=g(x), which in turn make du=f(x)dxdu=f(x)dx and dv=g(x)dx,dv=g(x)dx, we have the more compact form

udv=uvvdu.udv=uvvdu.
Theorem 3.1

Integration by Parts

Let u=f(x)u=f(x) and v=g(x)v=g(x) be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:

udv=uvvdu.udv=uvvdu.
3.1

The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.

Example 3.1

Using Integration by Parts

Use integration by parts with u=xu=x and dv=sinxdxdv=sinxdx to evaluate xsinxdx.xsinxdx.

Solution

By choosing u=x,u=x, we have du=1dx.du=1dx. Since dv=sinxdx,dv=sinxdx, we get v=sinxdx=cosx.v=sinxdx=cosx. It is handy to keep track of these values as follows:

u=xdv=sinxdxdu=1dxv=sinxdx=cosx.u=xdv=sinxdxdu=1dxv=sinxdx=cosx.

Applying the integration-by-parts formula results in

xsinxdx=(x)(cosx)(cosx)(1dx)Substitute.=xcosx+cosxdxSimplify.=xcosx+sinx+C.Usecosxdx=sinx+C.xsinxdx=(x)(cosx)(cosx)(1dx)Substitute.=xcosx+cosxdxSimplify.=xcosx+sinx+C.Usecosxdx=sinx+C.

Analysis

At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen u=sinxu=sinx and dv=x.dv=x. If we had done so, then we would have du=cosxdu=cosx and v=12x2.v=12x2. Thus, after applying integration by parts, we have xsinxdx=12x2sinx12x2cosxdx.xsinxdx=12x2sinx12x2cosxdx. Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for uu and dvdv before finding a choice that works.

Second, you may wonder why, when we find v=sinxdx=cosx,v=sinxdx=cosx, we do not use v=cosx+K.v=cosx+K. To see that it makes no difference, we can rework the problem using v=cosx+K:v=cosx+K:

xsinxdx=(x)(cosx+K)(cosx+K)(1dx)=xcosx+Kx+cosxdxKdx=xcosx+Kx+sinxKx+C=xcosx+sinx+C.xsinxdx=(x)(cosx+K)(cosx+K)(1dx)=xcosx+Kx+cosxdxKdx=xcosx+Kx+sinxKx+C=xcosx+sinx+C.

As you can see, it makes no difference in the final solution.

Last, we can check to make sure that our antiderivative is correct by differentiating xcosx+sinx+C:xcosx+sinx+C:

ddx(xcosx+sinx+C)=(−1)cosx+(x)(sinx)+cosx=xsinx.ddx(xcosx+sinx+C)=(−1)cosx+(x)(sinx)+cosx=xsinx.

Therefore, the antiderivative checks out.

Media

Watch this video and visit this website for examples of integration by parts.

Checkpoint 3.1

Evaluate xe2xdxxe2xdx using the integration-by-parts formula with u=xu=x and dv=e2xdx.dv=e2xdx.

The natural question to ask at this point is: How do we know how to choose uu and dv?dv? Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. This mnemonic serves as an aid in determining an appropriate choice for u.u.

The type of function in the integral that appears first in the list should be our first choice of u.u. For example, if an integral contains a logarithmic function and an algebraic function, we should choose uu to be the logarithmic function, because L comes before A in LIATE. The integral in Example 3.1 has a trigonometric function (sinx)(sinx) and an algebraic function (x).(x). Because A comes before T in LIATE, we chose uu to be the algebraic function. When we have chosen u,u, dvdv is selected to be the remaining part of the function to be integrated, together with dx.dx.

Why does this mnemonic work? Remember that whatever we pick to be dvdv must be something we can integrate. Since we do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions, it makes sense that they should not be chosen as values for dv.dv. Consequently, they should be at the head of the list as choices for u.u. Thus, we put LI at the beginning of the mnemonic. (We could just as easily have started with IL, since these two types of functions won’t appear together in an integration-by-parts problem.) The exponential and trigonometric functions are at the end of our list because they are fairly easy to integrate and make good choices for dv.dv. Thus, we have TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appear together it usually doesn’t really matter which one is uu and which one is dv.)dv.) Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.

Example 3.2

Using Integration by Parts

Evaluate lnxx3dx.lnxx3dx.

Solution

Begin by rewriting the integral:

lnxx3dx=x−3lnxdx.lnxx3dx=x−3lnxdx.

Since this integral contains the algebraic function x−3x−3 and the logarithmic function lnx,lnx, choose u=lnx,u=lnx, since L comes before A in LIATE. After we have chosen u=lnx,u=lnx, we must choose dv=x−3dx.dv=x−3dx.

Next, since u=lnx,u=lnx, we have du=1xdx.du=1xdx. Also, v=x−3dx=12x−2.v=x−3dx=12x−2. Summarizing,

u=lnxdv=x−3dxdu=1xdxv=x−3dx=12x−2.u=lnxdv=x−3dxdu=1xdxv=x−3dx=12x−2.

Substituting into the integration-by-parts formula (Equation 3.1) gives

lnxx3dx=x−3lnxdx=(lnx)(12x−2)(12x−2)(1xdx)=12x−2lnx+12x−3dxSimplify.=12x−2lnx14x−2+CIntegrate.=12x2lnx14x2+C.Rewrite with positive integers.lnxx3dx=x−3lnxdx=(lnx)(12x−2)(12x−2)(1xdx)=12x−2lnx+12x−3dxSimplify.=12x−2lnx14x−2+CIntegrate.=12x2lnx14x2+C.Rewrite with positive integers.
Checkpoint 3.2

Evaluate xlnxdx.xlnxdx.

In some cases, as in the next two examples, it may be necessary to apply integration by parts more than once.

Example 3.3

Applying Integration by Parts More Than Once

Evaluate x2e3xdx.x2e3xdx.

Solution

Using LIATE, choose u=x2u=x2 and dv=e3xdx.dv=e3xdx. Thus, du=2xdxdu=2xdx and v=e3xdx=(13)e3x.v=e3xdx=(13)e3x. Therefore,

u=x2dv=e3xdxdu=2xdxv=e3xdx=13e3x.u=x2dv=e3xdxdu=2xdxv=e3xdx=13e3x.

Substituting into Equation 3.1 produces

x2e3xdx=13x2e3x23xe3xdx.x2e3xdx=13x2e3x23xe3xdx.

We still cannot integrate 23xe3xdx23xe3xdx directly, but the integral now has a lower power on x.x. We can evaluate this new integral by using integration by parts again. To do this, choose u=xu=x and dv=23e3xdx.dv=23e3xdx. Thus, du=dxdu=dx and v=(23)e3xdx=(29)e3x.v=(23)e3xdx=(29)e3x. Now we have

u=xdv=23e3xdxdu=dxv=23e3xdx=29e3x.u=xdv=23e3xdxdu=dxv=23e3xdx=29e3x.

Substituting back into the previous equation yields

x2e3xdx=13x2e3x(29xe3x29e3xdx).x2e3xdx=13x2e3x(29xe3x29e3xdx).

After evaluating the last integral and simplifying, we obtain

x2e3xdx=13x2e3x29xe3x+227e3x+C.x2e3xdx=13x2e3x29xe3x+227e3x+C.

Example 3.4

Applying Integration by Parts When LIATE Doesn’t Quite Work

Evaluate t3et2dt.t3et2dt.

Solution

If we use a strict interpretation of the mnemonic LIATE to make our choice of u,u, we end up with u=t3u=t3 and dv=et2dt.dv=et2dt. Unfortunately, this choice won’t work because we are unable to evaluate et2dt.et2dt. However, since we can evaluate tet2dx,tet2dx, we can try choosing u=t2u=t2 and dv=tet2dt.dv=tet2dt. With these choices we have

u=t2dv=tet2dtdu=2tdtv=tet2dt=12et2.u=t2dv=tet2dtdu=2tdtv=tet2dt=12et2.

Thus, we obtain

t3et2dt=12t2et212et22tdt=12t2et212et2+C.t3et2dt=12t2et212et22tdt=12t2et212et2+C.

Example 3.5

Applying Integration by Parts More Than Once

Evaluate sin(lnx)dx.sin(lnx)dx.

Solution

This integral appears to have only one function—namely, sin(lnx)sin(lnx) —however, we can always use the constant function 1 as the other function. In this example, let’s choose u=sin(lnx)u=sin(lnx) and dv=1dx.dv=1dx. (The decision to use u=sin(lnx)u=sin(lnx) is easy. We can’t choose dv=sin(lnx)dxdv=sin(lnx)dx because if we could integrate it, we wouldn’t be using integration by parts in the first place!) Consequently, du=(1/x)cos(lnx)dxdu=(1/x)cos(lnx)dx and v=1dx=x.v=1dx=x. After applying integration by parts to the integral and simplifying, we have

sin(lnx)dx=xsin(lnx)cos(lnx)dx.sin(lnx)dx=xsin(lnx)cos(lnx)dx.

Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let’s see what happens when we apply integration by parts again. This time let’s choose u=cos(lnx)u=cos(lnx) and dv=1dx,dv=1dx, making du=(1/x)sin(lnx)dxdu=(1/x)sin(lnx)dx and v=1dx=x.v=1dx=x. Substituting, we have

sin(lnx)dx=xsin(lnx)(xcos(lnx)sin(lnx)dx).sin(lnx)dx=xsin(lnx)(xcos(lnx)sin(lnx)dx).

After simplifying, we obtain

sin(lnx)dx=xsin(lnx)xcos(lnx)sin(lnx)dx.sin(lnx)dx=xsin(lnx)xcos(lnx)sin(lnx)dx.

The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, substitute I=sin(lnx)dx.I=sin(lnx)dx. Thus, the equation becomes

I=xsin(lnx)xcos(lnx)I.I=xsin(lnx)xcos(lnx)I.

First, add II to both sides of the equation to obtain

2I=xsin(lnx)xcos(lnx).2I=xsin(lnx)xcos(lnx).

Next, divide by 2:

I=12xsin(lnx)12xcos(lnx).I=12xsin(lnx)12xcos(lnx).

Substituting I=sin(lnx)dxI=sin(lnx)dx again, we have

sin(lnx)dx=12xsin(lnx)12xcos(lnx).sin(lnx)dx=12xsin(lnx)12xcos(lnx).

From this we see that (1/2)xsin(lnx)(1/2)xcos(lnx)(1/2)xsin(lnx)(1/2)xcos(lnx) is an antiderivative of sin(lnx)dx.sin(lnx)dx. For the most general antiderivative, add +C:+C:

sin(lnx)dx=12xsin(lnx)12xcos(lnx)+C.sin(lnx)dx=12xsin(lnx)12xcos(lnx)+C.

Analysis

If this method feels a little strange at first, we can check the answer by differentiation:

ddx(12xsin(lnx)12xcos(lnx))=12(sin(lnx))+cos(lnx)·1x·12x(12cos(lnx)sin(lnx)·1x·12x)=sin(lnx).ddx(12xsin(lnx)12xcos(lnx))=12(sin(lnx))+cos(lnx)·1x·12x(12cos(lnx)sin(lnx)·1x·12x)=sin(lnx).
Checkpoint 3.3

Evaluate x2sinxdx.x2sinxdx.

Integration by Parts for Definite Integrals

Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

Theorem 3.2

Integration by Parts for Definite Integrals

Let u=f(x)u=f(x) and v=g(x)v=g(x) be functions with continuous derivatives on [a,b].[a,b]. Then

abudv=uv|ababvdu.abudv=uv|ababvdu.
3.2

Example 3.6

Finding the Area of a Region

Find the area of the region bounded above by the graph of y=tan−1xy=tan−1x and below by the xx-axis over the interval [0,1].[0,1].

Solution

This region is shown in Figure 3.2. To find the area, we must evaluate 01tan−1xdx.01tan−1xdx.

This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.
Figure 3.2 To find the area of the shaded region, we have to use integration by parts.

For this integral, let’s choose u=tan−1xu=tan−1x and dv=dx,dv=dx, thereby making du=1x2+1dxdu=1x2+1dx and v=x.v=x. After applying the integration-by-parts formula (Equation 3.2) we obtain

Area=xtan−1x|0101xx2+1dx.Area=xtan−1x|0101xx2+1dx.

Use u-substitution to obtain

01xx2+1dx=12ln|x2+1|01.01xx2+1dx=12ln|x2+1|01.

Thus,

Area=xtan1x|0112ln|x2+1||01=π412ln2.Area=xtan1x|0112ln|x2+1||01=π412ln2.

At this point it might not be a bad idea to do a “reality check” on the reasonableness of our solution. Since π412ln20.4388,π412ln20.4388, and from Figure 3.2 we expect our area to be slightly less than 0.5, this solution appears to be reasonable.

Example 3.7

Finding a Volume of Revolution

Find the volume of the solid obtained by revolving the region bounded by the graph of f(x)=ex,f(x)=ex, the x-axis, the y-axis, and the line x=1x=1 about the y-axis.

Solution

The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).

This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.
Figure 3.3 We can use the shell method to find a volume of revolution.

To find the volume using shells, we must evaluate 2π01xexdx.2π01xexdx. To do this, let u=xu=x and dv=ex.dv=ex. These choices lead to du=dxdu=dx and v=ex=ex.v=ex=ex. Substituting into Equation 3.2, we obtain

Volume=2π01xexdx=2π(xex|01+01exdx)Use integration by parts.=−2πxex|012πex|01Evaluate01exdx=ex|01.=2π4πe.Evaluate and simplify.Volume=2π01xexdx=2π(xex|01+01exdx)Use integration by parts.=−2πxex|012πex|01Evaluate01exdx=ex|01.=2π4πe.Evaluate and simplify.

Analysis

Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius 11 and height of 1/e1/e added to the volume of a cone of base radius 11 and height of 113.113. Consequently, the solid should have a volume a bit less than

π(1)21e+(π3)(1)2(11e)=2π3eπ31.8177.π(1)21e+(π3)(1)2(11e)=2π3eπ31.8177.

Since 2π4πe1.6603,2π4πe1.6603, we see that our calculated volume is reasonable.

Checkpoint 3.4

Evaluate 0π/2xcosxdx.0π/2xcosxdx.

Section 3.1 Exercises

In using the technique of integration by parts, you must carefully choose which expression is u. For each of the following problems, use the guidelines in this section to choose u. Do not evaluate the integrals.

1.

x3e2xdxx3e2xdx

2.

x3ln(x)dxx3ln(x)dx

3.

y3cosydxy3cosydx

4.

x2arctanxdxx2arctanxdx

5.

e3xsin(2x)dxe3xsin(2x)dx

Find the integral by using the simplest method. Not all problems require integration by parts.

6.

vsinvdvvsinvdv

7.

lnxdxlnxdx (Hint: lnxdxlnxdx is equivalent to 1·ln(x)dx.)1·ln(x)dx.)

8.

xcosxdxxcosxdx

9.

tan−1xdxtan−1xdx

10.

x2exdxx2exdx

11.

xsin(2x)dxxsin(2x)dx

12.

xe4xdxxe4xdx

13.

xexdxxexdx

14.

xcos3xdxxcos3xdx

15.

x2cosxdxx2cosxdx

16.

xlnxdxxlnxdx

17.

ln(2x+1)dxln(2x+1)dx

18.

x2e4xdxx2e4xdx

19.

exsinxdxexsinxdx

20.

excosxdxexcosxdx

21.

xex2dxxex2dx

22.

x2exdxx2exdx

23.

sin(ln(2x))dxsin(ln(2x))dx

24.

cos(lnx)dxcos(lnx)dx

25.

(lnx)2dx(lnx)2dx

26.

ln(x2)dxln(x2)dx

27.

x2lnxdxx2lnxdx

28.

sin−1xdxsin−1xdx

29.

cos−1(2x)dxcos−1(2x)dx

30.

xarctanxdxxarctanxdx

31.

x2sinxdxx2sinxdx

32.

x3cosxdxx3cosxdx

33.

x3sinxdxx3sinxdx

34.

x3exdxx3exdx

35.

xsec−1xdxxsec−1xdx

36.

xsec2xdxxsec2xdx

37.

xcoshxdxxcoshxdx

Compute the definite integrals. Use a graphing utility to confirm your answers.

38.

1/e1lnxdx1/e1lnxdx

39.

01xe−2xdx01xe−2xdx (Express the answer in exact form.)

40.

01exdx(letu=x)01exdx(letu=x)

41.

1eln(x2)dx1eln(x2)dx

42.

0πxcosxdx0πxcosxdx

43.

ππxsinxdxππxsinxdx (Express the answer in exact form.)

44.

03ln(x2+1)dx03ln(x2+1)dx (Express the answer in exact form.)

45.

0π/2x2sinxdx0π/2x2sinxdx (Express the answer in exact form.)

46.

01x5xdx01x5xdx (Express the answer using five significant digits.)

47.

Evaluate cosxln(sinx)dxcosxln(sinx)dx

Derive the following formulas using the technique of integration by parts. Assume that n is a positive integer. These formulas are called reduction formulas because the exponent in the x term has been reduced by one in each case. The second integral is simpler than the original integral.

48.

xnexdx=xnexnxn1exdxxnexdx=xnexnxn1exdx

49.

xncosxdx=xnsinxnxn1sinxdxxncosxdx=xnsinxnxn1sinxdx

50.

xnsinxdx=______xnsinxdx=______

51.

Integrate 2x2x3dx2x2x3dx using two methods:

  1. Using parts, letting dv=2x3dxdv=2x3dx
  2. Substitution, letting u=2x3u=2x3

State whether you would use integration by parts to evaluate the integral. If so, identify u and dv. If not, describe the technique used to perform the integration without actually doing the problem.

52.

xlnxdxxlnxdx

53.

ln2xxdxln2xxdx

54.

xexdxxexdx

55.

xex23dxxex23dx

56.

x2sinxdxx2sinxdx

57.

x2sin(3x3+2)dxx2sin(3x3+2)dx

Sketch the region bounded above by the curve, the x-axis, and x=1,x=1, and find the area of the region. Provide the exact form or round answers to the number of places indicated.

58.

y=2xexy=2xex (Approximate answer to four decimal places.)

59.

y=exsin(πx)y=exsin(πx) (Approximate answer to five decimal places.)

Find the volume generated by rotating the region bounded by the given curves about the specified line. Express the answers in exact form or approximate to the number of decimal places indicated.

60.

y=sinx,y=0,x=2π,x=3πy=sinx,y=0,x=2π,x=3π about the y-axis (Express the answer in exact form.)

61.

y=exy=ex y=0,x=−1x=0;y=0,x=−1x=0; about x=1x=1 (Express the answer in exact form.)

62.

A particle moving along a straight line has a velocity of v(t)=t2etv(t)=t2et after t sec. How far does it travel in the first 2 sec? (Assume the units are in feet and express the answer in exact form.)

63.

Find the area under the graph of y=sec3xy=sec3x from x=0tox=1.x=0tox=1. (Round the answer to two significant digits.)

64.

Find the area between y=(x2)exy=(x2)ex and the x-axis from x=2x=2 to x=5.x=5. (Express the answer in exact form.)

65.

Find the area of the region enclosed by the curve y=xcosxy=xcosx and the x-axis for

11π2x13π2.11π2x13π2. (Express the answer in exact form.)

66.

Find the volume of the solid generated by revolving the region bounded by the curve y=lnx,y=lnx, the x-axis, and the vertical line x=e2x=e2 about the x-axis. (Express the answer in exact form.)

67.

Find the volume of the solid generated by revolving the region bounded by the curve y=4cosxy=4cosx and the x-axis, π2x3π2,π2x3π2, about the x-axis. (Express the answer in exact form.)

68.

Find the volume of the solid generated by revolving the region in the first quadrant bounded by y=exy=ex and the x-axis, from x=0x=0 to x=ln(7),x=ln(7), about the y-axis. (Express the answer in exact form.)

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