Improper Integrals Comparison Quiz - Convergence, Comparison & Classic Forms

1) An improper integral can have:

A finite limit of integration and a continuous integrand.

An infinite limit of integration or a discontinuous integrand.

Only an infinite lower limit.

Only a discontinuous integrand at an endpoint.

By definition, an improper integral is one that does not satisfy the conditions of a standard definite integral because either the interval of integration is unbounded (one or both limits are infinite) or the integrand function has a discontinuity (becomes infinite) at some point in the interval. Therefore, the correct answer is that it has an infinite limit of integration or a discontinuous (unbounded) integrand.

Explanation

By definition, an improper integral is one that does not satisfy the conditions of a standard definite integral because either the interval of integration is unbounded (one or both limits are infinite) or the integrand function has a discontinuity (becomes infinite) at some point in the interval. Therefore, the correct answer is that it has an infinite limit of integration or a discontinuous (unbounded) integrand.

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

1/3

3

0

Diverges

This integral is improper due to the infinite upper limit. Set up: limit as b→∞ of ∫ from 0 to b of e^(-3x) dx. The antiderivative of e^(-3x) is (-1/3)e^(-3x). Evaluate from 0 to b: [(-1/3)e^(-3x)] from 0 to b = (-1/3)e^(-3b) - (-1/3)e^(0) = (-1/3)e^(-3b) + 1/3. Now take the limit as b→∞. As b→∞, e^(-3b) → 0. So the limit is 0 + 1/3 = 1/3. The integral converges to 1/3.

Explanation

This integral is improper due to the infinite upper limit. Set up: limit as b→∞ of ∫ from 0 to b of e^(-3x) dx. The antiderivative of e^(-3x) is (-1/3)e^(-3x). Evaluate from 0 to b: [(-1/3)e^(-3x)] from 0 to b = (-1/3)e^(-3b) - (-1/3)e^(0) = (-1/3)e^(-3b) + 1/3. Now take the limit as b→∞. As b→∞, e^(-3b) → 0. So the limit is 0 + 1/3 = 1/3. The integral converges to 1/3.

3) The integral ∫₀⁴ (1/(x-2)²) dx is improper because

The integrand is undefined at x=2, which is inside the interval.

The upper limit is 4.

The integrand is undefined at x=0.

The lower limit is 0.

The integrand 1/(x-2)² has a denominator that becomes zero when x=2. At x=2, the function is undefined and its limit is infinity. Since x=2 lies within the interval from 0 to 4, the integrand is unbounded at an interior point. This makes the integral improper.

Explanation

The integrand 1/(x-2)² has a denominator that becomes zero when x=2. At x=2, the function is undefined and its limit is infinity. Since x=2 lies within the interval from 0 to 4, the integrand is unbounded at an interior point. This makes the integral improper.

4) Evaluate ∫ from 1 to ∞ of (1/x⁴) dx.

1/3

1/4

1/3

Diverges

Set up as a limit: limit as b→∞ of ∫ from 1 to b of (1/x⁴) dx = ∫ from 1 to b of x^(-4) dx. The antiderivative is (x^(-3))/(-3) = -1/(3x³). Evaluate from 1 to b: [-1/(3x³)] from 1 to b = (-1/(3b³)) - (-1/(3*1³)) = -1/(3b³) + 1/3. Take the limit as b→∞: limit of -1/(3b³) is 0, so the result is 1/3. Thus the integral converges to 1/3.

Explanation

Set up as a limit: limit as b→∞ of ∫ from 1 to b of (1/x⁴) dx = ∫ from 1 to b of x^(-4) dx. The antiderivative is (x^(-3))/(-3) = -1/(3x³). Evaluate from 1 to b: [-1/(3x³)] from 1 to b = (-1/(3b³)) - (-1/(3*1³)) = -1/(3b³) + 1/3. Take the limit as b→∞: limit of -1/(3b³) is 0, so the result is 1/3. Thus the integral converges to 1/3.

5) What is the value of ∫ from -∞ to ∞ of e^(-|x|) dx?

0

1/1

1/2

Diverges

The function e^(-|x|) is even. We can write the integral as 2 ∫ from 0 to ∞ of e⁻ˣ dx (since for x≥0, |x|=x). So we compute 2 * ∫ from 0 to ∞ of e⁻ˣ dx. We know ∫ from 0 to ∞ of e⁻ˣ dx = 1 (from earlier examples). Therefore, 2 * 1 = 2. Alternatively, do it directly: ∫ from -∞ to ∞ = ∫ from -∞ to 0 e^(-(-x)) dx + ∫ from 0 to ∞ e⁻ˣ dx = ∫ from -∞ to 0 eˣ dx + ∫ from 0 to ∞ e⁻ˣ dx. The first integral is limit as a→-∞ of ∫ from a to 0 eˣ dx = limit a→-∞ (1 - e^a) = 1. The second is 1. Sum = 2.

Explanation

The function e^(-|x|) is even. We can write the integral as 2 ∫ from 0 to ∞ of e⁻ˣ dx (since for x≥0, |x|=x). So we compute 2 * ∫ from 0 to ∞ of e⁻ˣ dx. We know ∫ from 0 to ∞ of e⁻ˣ dx = 1 (from earlier examples). Therefore, 2 * 1 = 2. Alternatively, do it directly: ∫ from -∞ to ∞ = ∫ from -∞ to 0 e^(-(-x)) dx + ∫ from 0 to ∞ e⁻ˣ dx = ∫ from -∞ to 0 eˣ dx + ∫ from 0 to ∞ e⁻ˣ dx. The first integral is limit as a→-∞ of ∫ from a to 0 eˣ dx = limit a→-∞ (1 - e^a) = 1. The second is 1. Sum = 2.

6) Determine if ∫₀¹ (1/√(x)) dx converges or diverges. If it converges, find its value.

Converges to 0

Converges to 2

Diverges

Converges to 1

The integrand is 1/√x = x^(-½), which is unbounded as x→0⁺. Set up: limit as a→0⁺ of ∫ from a to 1 of x^(-½) dx. The antiderivative is 2√x. Evaluate: [2√x] from a to 1 = 2√1 - 2√a = 2 - 2√a. Take limit as a→0⁺: 2 - 0 = 2. So it converges to 2.

Explanation

The integrand is 1/√x = x^(-½), which is unbounded as x→0⁺. Set up: limit as a→0⁺ of ∫ from a to 1 of x^(-½) dx. The antiderivative is 2√x. Evaluate: [2√x] from a to 1 = 2√1 - 2√a = 2 - 2√a. Take limit as a→0⁺: 2 - 0 = 2. So it converges to 2.

7) When evaluating ∫ from 1 to ∞ of (1/x^p) dx, divergence occurs when:

P > 1

P < 1

P = 1

Both B and C

The p-test states that ∫ from 1 to ∞ of (1/x^p) dx converges if p > 1 and diverges if p ≤ 1. Therefore, divergence occurs when p is less than 1 or equal to 1. So both options B (p

Explanation

The p-test states that ∫ from 1 to ∞ of (1/x^p) dx converges if p > 1 and diverges if p ≤ 1. Therefore, divergence occurs when p is less than 1 or equal to 1. So both options B (p

8) Evaluate ∫ from 0 to ∞ of (1/(x²+4)) dx.

π/4

π/2

π

Diverges

Set up: limit as b→∞ of ∫ from 0 to b of (1/(x²+4)) dx. Recall that ∫ (1/(x²+a²)) dx = (1/a) arctan(x/a). Here a²=4, so a=2. Thus antiderivative is (½) arctan(x/2). Evaluate from 0 to b: [(½) arctan(x/2)] from 0 to b = (½) arctan(b/2) - (½) arctan(0) = (½) arctan(b/2). Take limit as b→∞: arctan(b/2) → π/2, so (½)*(π/2) = π/4. The integral converges to π/4.

Explanation

Set up: limit as b→∞ of ∫ from 0 to b of (1/(x²+4)) dx. Recall that ∫ (1/(x²+a²)) dx = (1/a) arctan(x/a). Here a²=4, so a=2. Thus antiderivative is (½) arctan(x/2). Evaluate from 0 to b: [(½) arctan(x/2)] from 0 to b = (½) arctan(b/2) - (½) arctan(0) = (½) arctan(b/2). Take limit as b→∞: arctan(b/2) → π/2, so (½)*(π/2) = π/4. The integral converges to π/4.

9) The integral ∫₀¹ (1/x) dx is:

Convergent and equals 1.

Convergent and equals 0.

Divergent because the limit is infinite.

Divergent because the integrand is not defined at x=0.

Evaluate: limit as a→0⁺ of ∫ from a to 1 of (1/x) dx = limit a→0⁺ of [ln|x|] from a to 1 = ln(1) - ln(a) = 0 - ln(a). As a→0⁺, ln(a) → -∞, so -ln(a) → ∞. Therefore, the integral diverges to infinity. While it is true the integrand is not defined at x=0 (choice D), the precise reason for divergence is that the limit is infinite. Choice C is more specific to the outcome.

Explanation

Evaluate: limit as a→0⁺ of ∫ from a to 1 of (1/x) dx = limit a→0⁺ of [ln|x|] from a to 1 = ln(1) - ln(a) = 0 - ln(a). As a→0⁺, ln(a) → -∞, so -ln(a) → ∞. Therefore, the integral diverges to infinity. While it is true the integrand is not defined at x=0 (choice D), the precise reason for divergence is that the limit is infinite. Choice C is more specific to the outcome.

10) True or False: If the improper integral of |f(x)| from 1 to infinity converges, then the improper integral of f(x) from 1 to infinity must also converge.

True

False

This is the definition of Absolute Convergence. If the integral of the absolute value of a function converges, the function is said to converge absolutely. A theorem in calculus states that absolute convergence implies convergence. Intuitively, if the "total amount" of area (ignoring signs) is finite, then the net area (considering signs) must also be finite.

Explanation

This is the definition of Absolute Convergence. If the integral of the absolute value of a function converges, the function is said to converge absolutely. A theorem in calculus states that absolute convergence implies convergence. Intuitively, if the "total amount" of area (ignoring signs) is finite, then the net area (considering signs) must also be finite.

11) Which of the following statements is TRUE about improper integrals?

All improper integrals diverge.

If an improper integral converges, its value is always positive.

An improper integral may converge to a finite negative number.

An improper integral cannot have both an infinite limit and an unbounded integrand.

Check each: A is false; many improper integrals converge. B is false; for example, ∫₀¹ ln(x) dx converges to -1. C is true, as shown by the ln(x) example. D is false; an integral can have both features, like ∫ from 0 to ∞ of (1/√x) dx is improper at 0 (unbounded) and at ∞ (infinite limit). So C is correct.

Explanation

Check each: A is false; many improper integrals converge. B is false; for example, ∫₀¹ ln(x) dx converges to -1. C is true, as shown by the ln(x) example. D is false; an integral can have both features, like ∫ from 0 to ∞ of (1/√x) dx is improper at 0 (unbounded) and at ∞ (infinite limit). So C is correct.

12) Evaluate ∫₀¹ (1/√(1-x²)) dx.

π/2

π

1

Diverges

The integrand 1/√(1-x²) is unbounded as x→1⁻. So we set up: limit as b→1⁻ of ∫ from 0 to b of (1/√(1-x²)) dx. The antiderivative of 1/√(1-x²) is arcsin(x). Evaluate: [arcsin(x)] from 0 to b = arcsin(b) - arcsin(0) = arcsin(b). Take limit as b→1⁻: arcsin(b) → arcsin(1) = π/2. So the integral converges to π/2.

Explanation

The integrand 1/√(1-x²) is unbounded as x→1⁻. So we set up: limit as b→1⁻ of ∫ from 0 to b of (1/√(1-x²)) dx. The antiderivative of 1/√(1-x²) is arcsin(x). Evaluate: [arcsin(x)] from 0 to b = arcsin(b) - arcsin(0) = arcsin(b). Take limit as b→1⁻: arcsin(b) → arcsin(1) = π/2. So the integral converges to π/2.

13) The integral ∫ from 1 to ∞ of (1/x^0.99) dx:

Converges because 0.99 < 1.

Diverges because 0.99 < 1.

Converges because 0.99 > 0.

Diverges because it's a reciprocal function.

This is of the form ∫ from 1 to ∞ of 1/x^p dx with p=0.99. The p-test says that if p ≤ 1, the integral diverges. Since 0.99 1. So B is correct.

Explanation

This is of the form ∫ from 1 to ∞ of 1/x^p dx with p=0.99. The p-test says that if p ≤ 1, the integral diverges. Since 0.99 1. So B is correct.

14) Evaluate ∫ from 0 to ∞ of (x e^(-x²)) dx.

1/2

1

0

Diverges

Set up: limit as b→∞ of ∫ from 0 to b of x e^(-x²) dx. Use substitution u = x², du = 2x dx, so x dx = du/2. When x=0, u=0; when x=b, u=b². The integral becomes ∫ from u=0 to u=b² of e^(-u) * (du/2) = (½) ∫ from 0 to b² of e^(-u) du = (½)[-e^(-u)] from 0 to b² = (½)(-e^(-b²) + e^0) = (½)(1 - e^(-b²)). Now take limit as b→∞: e^(-b²) → 0, so limit = (½)(1 - 0) = 1/2. Thus converges to 1/2.

Explanation

Set up: limit as b→∞ of ∫ from 0 to b of x e^(-x²) dx. Use substitution u = x², du = 2x dx, so x dx = du/2. When x=0, u=0; when x=b, u=b². The integral becomes ∫ from u=0 to u=b² of e^(-u) * (du/2) = (½) ∫ from 0 to b² of e^(-u) du = (½)[-e^(-u)] from 0 to b² = (½)(-e^(-b²) + e^0) = (½)(1 - e^(-b²)). Now take limit as b→∞: e^(-b²) → 0, so limit = (½)(1 - 0) = 1/2. Thus converges to 1/2.