Math: Improper Integrals and Convergence Tests Worksheet

Question 1

Evaluate the improper integral ∫ from 1 to infinity of 1/x^2 dx, or state that it diverges.

Accepted Answer

The integral converges. Writing it as lim as b approaches infinity of ∫ from 1 to b x^-2 dx gives lim as b approaches infinity of [-1/x] from 1 to b, which is lim as b approaches infinity of 1 - 1/b = 1.

Question 2

Determine whether ∫ from 1 to infinity of 1/x dx converges or diverges. Explain your reasoning.

Accepted Answer

The integral diverges. Writing it as lim as b approaches infinity of ∫ from 1 to b 1/x dx gives lim as b approaches infinity of ln b - ln 1, which grows without bound.

Question 3

Evaluate the improper integral ∫ from 0 to 1 of 1/sqrt(x) dx, or state that it diverges.

Accepted Answer

The integral converges. Writing it as lim as a approaches 0 from the right of ∫ from a to 1 x^-1/2 dx gives lim as a approaches 0 from the right of [2sqrt(x)] from a to 1, which equals 2.

Question 4

Determine whether ∫ from 0 to 1 of 1/x^2 dx converges or diverges. Explain your reasoning.

Accepted Answer

The integral diverges. Writing it as lim as a approaches 0 from the right of ∫ from a to 1 x^-2 dx gives lim as a approaches 0 from the right of [-1/x] from a to 1 = lim as a approaches 0 from the right of -1 + 1/a, which grows without bound.

Question 5

Use the p-integral test to determine whether ∫ from 1 to infinity of 1/x^p dx converges when p = 3/2. Do not compute the exact value unless needed.

Accepted Answer

The integral converges because it is a p-integral on [1, infinity) with p = 3/2, and p-integrals of the form ∫ from 1 to infinity 1/x^p dx converge when p > 1.

Question 6

Use the p-integral test to determine whether ∫ from 1 to infinity of 1/x^0.8 dx converges or diverges.

Accepted Answer

The integral diverges because it is a p-integral on [1, infinity) with p = 0.8, and p-integrals of the form ∫ from 1 to infinity 1/x^p dx diverge when p is less than or equal to 1.

Question 7

Determine whether ∫ from 2 to infinity of 5/(x^2 + 1) dx converges or diverges. Use comparison with a simpler function.

Accepted Answer

The integral converges. For x greater than or equal to 2, x^2 + 1 is greater than x^2, so 5/(x^2 + 1) is less than 5/x^2. Since ∫ from 2 to infinity 5/x^2 dx converges, the given integral converges by direct comparison.

Question 8

Determine whether ∫ from 1 to infinity of x/(x^2 + 1) dx converges or diverges.

Accepted Answer

The integral diverges. An antiderivative is (1/2)ln(x^2 + 1), so the improper integral becomes lim as b approaches infinity of (1/2)ln(b^2 + 1) - (1/2)ln 2, which grows without bound.

Question 9

Use limit comparison to determine whether ∫ from 1 to infinity of (3x + 2)/(x^2 + 4) dx converges or diverges.

Accepted Answer

The integral diverges. Compare the integrand to 1/x. The limit as x approaches infinity of [(3x + 2)/(x^2 + 4)] divided by [1/x] is the limit of x(3x + 2)/(x^2 + 4), which equals 3. Since ∫ from 1 to infinity 1/x dx diverges and the limit is a positive finite number, the given integral diverges.

Question 10

Evaluate ∫ from negative infinity to 0 of e^x dx, or state that it diverges.

Accepted Answer

The integral converges. Writing it as lim as a approaches negative infinity of ∫ from a to 0 e^x dx gives lim as a approaches negative infinity of [e^x] from a to 0, which is 1 - 0 = 1.

Question 11

Determine whether ∫ from 0 to infinity of e^-x dx converges, and find its value if it converges.

Accepted Answer

The integral converges. Writing it as lim as b approaches infinity of ∫ from 0 to b e^-x dx gives lim as b approaches infinity of [-e^-x] from 0 to b, which is lim as b approaches infinity of 1 - e^-b = 1.

Question 12

An improper integral is written as ∫ from 0 to 3 of 1/(x - 2)^2 dx. Explain why it must be split before testing convergence, then determine whether it converges or diverges.

Accepted Answer

The integral must be split because the integrand has a vertical asymptote at x = 2, which lies inside the interval from 0 to 3. It must be considered as ∫ from 0 to 2 of 1/(x - 2)^2 dx plus ∫ from 2 to 3 of 1/(x - 2)^2 dx. At least one side diverges because the function behaves like 1/u^2 near u = 0, so the original improper integral diverges.

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Math: Improper Integrals and Convergence Tests

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