Improper Integrals – Convergence, Comparison & Classic Forms

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

The integrand is unbounded at x=1. For integrals of the form ∫ from a to b of (1/(x-c)^p) dx with singularity at c inside, convergence occurs if p