Calculus: Integrals (Advanced) Worksheet

Question 1

Evaluate the indefinite integral ∫ x^2 e^(3x) dx.

Accepted Answer

The integral is e^(3x)(x^2/3 - 2x/9 + 2/27) + C. This follows by applying integration by parts twice, or by using the repeated integration by parts pattern for a polynomial times an exponential function.

Question 2

Evaluate the definite integral ∫ from 0 to 1 of ln(1 + x) dx.

Accepted Answer

The value of the integral is 2ln(2) - 1. One way to compute it is to use integration by parts with u = ln(1 + x) and dv = dx.

Question 3

Evaluate ∫ dx/(x^2 sqrt(x^2 - 9)) for x > 3.

Accepted Answer

The integral is sqrt(x^2 - 9)/(9x) + C. Using the substitution x = 3sec(theta) changes the integral into a multiple of cos(theta), and converting back gives sqrt(x^2 - 9)/(9x) + C.

Question 4

Determine whether the improper integral ∫ from 1 to infinity of 1/(x(ln x)^2) dx converges or diverges. If it converges, find its value.

Accepted Answer

The integral converges and its value is 1. With u = ln x, du = dx/x, the integral becomes ∫ from 0 to infinity of 1/u^2 du, but the lower endpoint u = 0 is improper. Since the original lower bound x = 1 makes ln x = 0, the integral actually diverges at x = 1. Therefore the integral diverges.

Question 5

Evaluate the improper integral ∫ from e to infinity of 1/(x(ln x)^2) dx.

Accepted Answer

The integral converges and equals 1. With u = ln x, du = dx/x, the integral becomes ∫ from 1 to infinity of u^(-2) du, which equals 1.

Question 6

Evaluate ∫ from 0 to pi/2 of sin^5(x) cos^2(x) dx.

Accepted Answer

The value of the integral is 8/105. Writing sin^5(x) as sin(x)(1 - cos^2(x))^2 and using u = cos x gives ∫ from 0 to 1 of (1 - 2u^2 + u^4)u^2 du, which simplifies to 8/105.

Question 7

Find the area enclosed by y = x^2 and y = 2x from x = 0 to x = 2.

Accepted Answer

The enclosed area is 4/3 square units. On 0 ≤ x ≤ 2, the line y = 2x lies above the parabola y = x^2, so the area is ∫ from 0 to 2 of (2x - x^2) dx = 4/3.

Question 8

Compute the volume obtained by rotating the region bounded by y = sqrt(x), y = 0, and x = 4 about the x-axis.

Accepted Answer

The volume is 8pi cubic units. Using washers, the radius is sqrt(x), so V = pi ∫ from 0 to 4 of (sqrt(x))^2 dx = pi ∫ from 0 to 4 of x dx = 8pi.

Question 9

Evaluate ∫ from 0 to infinity of e^(-2x) cos(3x) dx.

Accepted Answer

The value of the integral is 2/13. Using the standard Laplace integral ∫ from 0 to infinity of e^(-ax)cos(bx) dx = a/(a^2 + b^2) for a > 0 gives 2/(4 + 9) = 2/13.

Question 10

Evaluate the integral ∫ x/(x^2 + 4x + 13) dx.

Accepted Answer

The integral is (1/2)ln(x^2 + 4x + 13) - (2/3)arctan((x + 2)/3) + C. Rewrite x as (1/2)(2x + 4) - 2, then integrate the logarithmic part and the arctangent part after completing the square.

Question 11

Use partial fractions to evaluate ∫ (3x + 5)/(x^2 - x - 2) dx.

Accepted Answer

The integral is (11/3)ln|x - 2| - (2/3)ln|x + 1| + C. Since x^2 - x - 2 = (x - 2)(x + 1), decomposing gives (3x + 5)/((x - 2)(x + 1)) = (11/3)/(x - 2) - (2/3)/(x + 1).

Question 12

Evaluate ∫ from 0 to 1 of x^3 sqrt(1 - x^2) dx.

Accepted Answer

The value of the integral is 2/15. Let u = 1 - x^2, so x dx = -du/2 and x^2 = 1 - u. The integral becomes (1/2)∫ from 0 to 1 of (1 - u)u^(1/2) du = 2/15.

Question 13

Find d/dx of F(x) = ∫ from x^2 to sin x of e^(t^3) dt.

Accepted Answer

The derivative is F'(x) = e^((sin x)^3)cos x - 2x e^(x^6). This follows from the Fundamental Theorem of Calculus with variable upper and lower limits.

Question 14

Evaluate ∫ from 0 to 1 of ∫ from y to 1 of e^(x^2) dx dy by reversing the order of integration.

Accepted Answer

The value is (e - 1)/2. The region is 0 ≤ y ≤ x ≤ 1, so reversing the order gives ∫ from 0 to 1 of ∫ from 0 to x of e^(x^2) dy dx = ∫ from 0 to 1 of x e^(x^2) dx = (e - 1)/2.

Question 15

Let I(a) = ∫ from 0 to infinity of e^(-ax) sin x dx for a > 0. Find I(a).

Accepted Answer

The function is I(a) = 1/(a^2 + 1). This is the standard Laplace transform of sin x, or it can be found by integrating e^(-ax)sin x by parts twice and using the convergence condition a > 0.

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Calculus: Integrals (Advanced)

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