APPENDIX
B

Integration Practice Problems and Solutions

Here’s your chance to go through a series of integration problems without knowing the chapter topic (which can hint at how to solve the problem). In this appendix, I provide a set of problems. Once you’ve worked through them, you can check your work against the solutions provided later in this appendix.

Good luck, and have fun with what you’ve learned!

Problems

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Consider the graph of f(x) = 2x² and the y-axis on the closed interval [1,4]. Use this information to answer Problems 15 through 17.

15. Determine the length of the arc of the curve in this interval. Let R be the region bounded by f(x) and the x-axis on the same interval. A solid is formed by rotating R around the x-axis.

16. Find the volume, V, of the solid.

17. Find the surface area, S, of the solid. Let R be the first quadrant region bounded by the graphs of f(x) = sin(x), g(x) = cos(x), and the y-axis. Use this information for Problems 18 through 20.

18. A solid with base R has cross sections perpendicular to the x-axis in the shape of squares. Find the volume of the solid.

19. Let S be the solid formed when R is rotated around the y-axis. Find the volume of S.

20. Let T be the solid when R is rotated around the x-axis. Find the volume of T.

Solutions

Let’s see how you did.

1.

The derivative of tan^–1(x) is , so let u = tan^–1(x) and dx. Substitution gives .

2.

The derivative of 2sec(x) – 1 is sec(x) tan(x), so let u = 2sec(x) – 1 and du = sec(x) tan(x)dx. Substitution gives .

3.

Rewrite as . The derivative of x² + 1 is 2x, so the numerator of the first fraction is half the derivative of the denominator. This means and that the second part of the integrand is the formula for the antiderivative of tan^–1(x).

4.

The key to this problem is the radical in the denominator contains the difference of two squares. If you let u = ln(2x), then . Substitution gives .

5.

The polynomial is not the derivative of the exponent in the exponential function. Let’s use the Tabular Method to solve this problem.

Figure B.1

6.

Rewrite tan³(x) as tan(x)tan²(x) and then use the Pythagorean identity tan2(x) = sec²(x) – 1. becomes . With sec²(x) being the derivative of tan(x), the antiderivative of tan(x)sec²(x) is ½tan²(x). Earlier, we saw that the antiderivative of tan(x) is ln|sec(x)|.

7.

(cos(x) + sin(x))² = cos²(x) + 2cos(x) sin(x) + sin²(x). Since cos²(x) + sin²(x) = 1 and 2cos(x)sin(x) = sin(2x), the integrand is now cos(2x)(1 + sin(2x)), which expands to cos(2x) + cos(2x)sin(2x). The antiderivative for cos(2x) is ½sin(2x). The second piece of the integrand, cos(2x)sin(2x), offers you two options, each is correct:

• If u = cos(2x), then du = –2sin(2x)dx and the problem becomes , which translates back to .
• If u = sin(2x), then du = 2cos(2x)dx. The resulting integral would be .

(This is another example of how the constant of integration absorbs any extra constants that are floating around. . The extra is absorbed by C.)

8.

Use cos³(x) = cos²(x)cos(x) = cos(x)(1 – sin²(x)). The integrand is now sin⁶(x) cos(x) – sin⁸(x)cos(x).

9.

Rewrite x⁴ as (x²)². The integrand is the sum of two squares leading us to believe this is some type of tan^–1(x) problem. Factor 4 from the denominator, . Let , so that du = xdx. Substitution gives . Rewrite the result in terms of the original variable to get .

10.

The denominator has a common factor of . What you need to see here is that the denominator contains the sum of squares. Let then du = ½x^–1/2dx, so 2du = x^–1/2dx. Substitution gives . Working back to the original variable, .

11.

This one is a little tricky. The answer is not is the derivative of tan^–1(x) not the antiderivative.) Use integration by parts with u = tan^–1(x) and dv = dx. You get du = and v = x.

. As we saw in Problem 3, Consequently, .

12.

This looks like Problem 10, which might get you to thinking about factoring from the denominator. . Let so that du = ½x^–1/2 dx and 2 du = x^–1/2 dx. Substitution gives . Going back to the original variable, .

13.

The product of the polynomial and trigonometric function should lead us to use integration by parts. Let u = x and , so that du = dx and v = .

14.

The denominator of the integrand factors to (x – 2)(x² + 1) leads us to use partial fractions as the technique to solve this problem.

Multiply through by the common denominator to get:

2x – 1 = A(x² + 1) + (Bx + C)(x – 2)

Consider the graph of f(x) = 2x² and the y-axis on the closed interval [1,4]. Use this information to answer Problems 15 through 17.

15. Determine the length of the arc of the curve in this interval.

The derivative of f(x) = 4x, so the length of the arc is . The sum of the squares within the radical is an indication that trigonometric substitution is a viable option.

Figure B.2

Let and tan(θ) = 4x so that .

becomes .

Returning to the original variable,
. In more manageable (meaningful?) numbers, the arc length is 30.1724.

Let R be the region bounded by f(x) and the x-axis on the same interval. A solid is formed by rotating R about the x-axis.

16. Find the volume, V, of the solid.

A cross section of the solid formed is a disk and the radius of this disk will be f(x). Therefore, the volume of the solid formed is .

17. Find the surface area, S, of the solid.

The surface area of the solid is . Using Figure B.2, let sec(θ) = and so that .
Converting the integral to trigonometric functions,

. Return to the original
variable, .

Let R be the first quadrant region bounded by the graphs of f(x) = sin(x), g(x) = cos(x), and the y-axis. Use this information for Problems 18 through 20.

The region R is:

Figure B.3

18. A solid with base R has cross sections perpendicular to the x-axis in the shape of squares. Find the volume of the solid.

The length of a side of the square is cos(x) – sin(x). Therefore, the volume of the solid formed is .

19. Let S be the solid formed when R is rotated about the y-axis. Find the volume of S.

Because we are revolving R around a vertical line, let’s use cylindrical shells. The radius of each cylinder is x, and the height of the cylinder is cos(x) – sin(x). Therefore, the volume of S is .

Each of these terms can be integrated using integration by parts.

For x cos(x), let u = x and dv = cos(x) dx, so that du = dx and v = sin(x).

For xsin(x), let u = x and dv = sin(x) dx, so that du = dx and v = –cos(x).

Therefore, .

20. Let T be the solid when R is rotated around the x-axis. Find the surface area of T.

A cross section of T shows a disk with larger radius cos(x) and smaller radius sin(x). The volume of T is . Use the original trigonometric identity for cos(2x) = cos²(x) – sin²(x) to get
.