Improper Integrals – Infinite Limits & p-Test Practice

An improper integral due to an unbounded integrand occurs when the function being integrated has a vertical asymptote (goes to infinity) at some point within the limits of integration. In choice B, the integrand is 1/(x-2). At x=2, this function is undefined and its limit is infinity. Since x=2 is between 1 and 5, the integral is improper. Choice A is improper due to an infinite limit. Choice C is improper due to an infinite limit. Choice D is a proper integral because the integrand is defined and continuous on [0,10].

The integral has an infinite upper limit. We set it up as a limit: limit as b→∞ of ∫ from 0 to b of (1/(1+x²)) dx. The antiderivative of 1/(1+x²) is arctan(x). Evaluate from 0 to b: [arctan(x)] from 0 to b = arctan(b) - arctan(0) = arctan(b) - 0 = arctan(b). Now take the limit as b→∞: limit as b→∞ of arctan(b). The arctangent function approaches π/2 as its argument goes to infinity. Therefore, the limit is π/2. The integral converges to π/2.

This is an improper integral with an infinite upper limit. Rewrite as a limit: limit as b→∞ of ∫ from 1 to b of (1/x³) dx. First, rewrite 1/x³ as x^(-3). Its antiderivative is (x^(-2))/(-2) = -1/(2x²). Evaluate from 1 to b: [-1/(2x²)] from 1 to b = (-1/(2b²)) - (-1/(2*1²)) = -1/(2b²) + 1/2. Now take the limit as b→∞: limit as b→∞ of (1/2 - 1/(2b²)). As b gets very large, 1/(2b²) approaches 0. Therefore, the limit is 1/2 - 0 = 1/2. The integral converges to 1/2.

The integrand is 1/√(9-x²). This function becomes undefined when the denominator is zero, i.e., when 9 - x² = 0, so x = 3 or x = -3. Within the interval [0,3], the point x=3 is an endpoint and at x=3, the denominator √(9-9)=0, so the function goes to infinity. Therefore, the integrand is unbounded at the upper limit x=3, making the integral improper.

The lower limit is -∞, so we set up: limit as a→-∞ of ∫ from a to 0 of eˣ dx. The antiderivative of eˣ is eˣ. Evaluate from a to 0: [eˣ] from a to 0 = e⁰- e^a = 1 - e^a. Now take the limit as a→-∞: limit as a→-∞ of (1 - e^a). As a becomes very negative, e^a approaches 0. Therefore, the limit is 1 - 0 = 1. The integral converges to 1.

The integrand 1/√(1-x) is unbounded as x approaches 1 from the left (since denominator → 0). So we set up: limit as b→1⁻ of ∫ from 0 to b of (1/√(1-x)) dx. Rewrite as (1-x)^(-½). Let u = 1-x, then du = -dx, so dx = -du. When x=0, u=1; when x=b, u=1-b. The integral becomes ∫ from u=1 to u=1-b of u^(-½) * (-du) = -∫ from 1 to 1-b of u^(-½) du = ∫ from 1-b to 1 of u^(-½) du. The antiderivative of u^(-½) is 2u^(½). So we have [2√u] from 1-b to 1 = 2√1 - 2√(1-b) = 2 - 2√(1-b). Now take the limit as b→1⁻: limit as b→1⁻ of (2 - 2√(1-b)). As b approaches 1, (1-b) approaches 0, so √(1-b) approaches 0. Thus the limit is 2 - 0 = 2. The integral converges to 2.

The integral is improper because the upper limit is infinity. The standard first step for such an integral is to replace the infinite limit with a variable and express the integral as a limit of a proper integral. Specifically, we write it as the limit as b approaches infinity of the integral from 2 to b of the function. This allows us to work with a definite integral with finite bounds before taking the limit.

At x = 0, the denominator is 0, making the expression 0/0 undefined. Even though the limit as x approaches 0 is 1 (a finite number), the function does not have a value at the endpoint x = 0. In strict calculus definitions, any integral where the function is undefined at a limit of integration is improper, even if the discontinuity is "removable."

We check each option for infinite limits or unbounded integrands. Option A: At x=1, the integrand 1/(x-1) is undefined, so it's improper. Option B: Has an infinite upper limit, so improper. Option C: The integrand ln(x-3) is undefined for x ≤ 3; at the lower limit x=3, ln(0) is undefined, so improper. Option D: √x is defined and continuous on [0,1] (even at x=0, √0=0, it's bounded). Therefore, this is a proper integral.

The p-test for improper integrals states that the integral ∫ from 1 to ∞ of 1/x^p dx converges if the exponent p is greater than 1. If p is less than or equal to 1, the integral diverges. This is a standard result in calculus. Therefore, the correct condition is p > 1.

This integral is improper due to the infinite limit. Set up: limit as b→∞ of ∫ from 0 to b of x e⁻ˣ dx. Use integration by parts. Let u = x, dv = e⁻ˣ dx. Then du = dx, v = -e⁻ˣ. The integral becomes: uv - ∫ v du = -x e⁻ˣ - ∫ (-e⁻ˣ) dx = -x e⁻ˣ + ∫ e⁻ˣ dx = -x e⁻ˣ - e⁻ˣ = -e⁻ˣ(x+1). Evaluate from 0 to b: [-e⁻ˣ(x+1)] from 0 to b = [-e^(-b)(b+1)] - [-e^(0)(0+1)] = -e^(-b)(b+1) + 1. Now take the limit as b→∞. Consider the term e^(-b)(b+1). As b→∞, e^(-b) goes to 0 faster than (b+1) grows, so the product goes to 0. Thus, the limit is 0 + 1 = 1. The integral converges to 1.

The integrand 1/(x-1) has a vertical asymptote at x=1, which is inside the interval (0,2). We must split the integral into two: from 0 to 1 and from 1 to 2. Consider the integral from 1 to 2: ∫ from 1 to 2 of (1/(x-1)) dx. Rewrite as limit as c→1⁺ of ∫ from c to 2 of (1/(x-1)) dx. The antiderivative is ln|x-1|. Evaluate: [ln|x-1|] from c to 2 = ln|1| - ln|c-1| = 0 - ln(c-1). As c→1⁺, (c-1)→0⁺, so ln(c-1)→ -∞. Therefore, this part diverges to -∞. Similarly, the left part diverges. Thus the whole integral diverges.

We check each using known convergence tests. For A: ∫ from 1 to ∞ of 1/√x dx = ∫ from 1 to ∞ of x^(-½) dx. By the p-test with p=1/2 ≤ 1, it diverges. For B: ∫ from 1 to ∞ of 1/x dx diverges (p=1). For C: ∫ from 1 to ∞ of 1/x³ dx, p=3 >1, so it converges. For D: ∫₀¹ 1/x dx diverges (unbounded at 0). Therefore, only C converges.

Use partial fractions: 1/(x(x+1)) = 1/x - 1/(x+1). So the integral becomes ∫ from 1 to ∞ of (1/x - 1/(x+1)) dx. Set up as limit: limit as b→∞ of ∫ from 1 to b of (1/x - 1/(x+1)) dx. The antiderivative is ln|x| - ln|x+1| = ln|x/(x+1)|. Evaluate from 1 to b: [ln|x/(x+1)|] from 1 to b = ln|b/(b+1)| - ln|1/2| = ln(b/(b+1)) - ln(½). Now take limit as b→∞: limit of ln(b/(b+1)). As b→∞, b/(b+1) → 1, so ln(1)=0. Thus the limit is 0 - ln(½) = -ln(½) = ln(2). So the integral converges to ln(2).

For the integral ∫₀¹ (1/x^p) dx, the integrand is unbounded at x=0. We can evaluate it as a limit: limit as a→0⁺ of ∫ from a to 1 of x^(-p) dx. The antiderivative is (x^(1-p))/(1-p) if p ≠ 1. Evaluate: [x^(1-p)/(1-p)] from a to 1 = (1^(1-p)/(1-p)) - (a^(1-p)/(1-p)) = 1/(1-p) - a^(1-p)/(1-p). As a→0⁺, a^(1-p) tends to 0 if 1-p > 0 (i.e., p1). If p=1, the antiderivative is ln x, which diverges. So the integral converges when p