# Improper Integrals (BC only) > AP Calculus BC · Unit 6: Integration and Accumulation of Change > Source: https://www.owlsprep.com/study/ap-calculus-bc-u6-improper-integrals/ This guide covers improper integrals with infinite bounds, discontinuous integrands, convergence definitions, the p-test, and comparison test for AP Calculus BC. Master evaluation and convergence testing for this BC-only topic. **Prerequisites:** [Definite integral evaluation via the Fundamental Theorem of Calculus](https://www.owlsprep.com/study/ap-calculus-ab-u5-fundamental-theorem-calculus/); [Limit computation at finite and infinite points](https://www.owlsprep.com/study/ap-calculus-ab-u1-limits-continuity/); [Basic integration rules for common functions](https://www.owlsprep.com/study/ap-calculus-bc-u6-basic-integration-rules/) ## Learning objectives - Define improper integrals and distinguish between Type 1 (infinite bounds) and Type 2 (discontinuous integrand) - Evaluate improper integrals using limits of proper definite integrals - Apply the p-test and comparison test to determine convergence/divergence - Avoid common exam pitfalls when working with improper integrals ## What is an Improper Integral? This topic is exclusive to AP Calculus BC and does not appear on the AP Calculus AB exam. It accounts for approximately 6-8% of the total AP Calculus BC exam score, and appears in both multiple-choice and free-response sections, often combined with other topics like integration techniques or infinite series. **Improper Integral** — A definite integral that fails at least one requirement for a proper definite integral: either the interval of integration is infinite, or the integrand has an infinite discontinuity (vertical asymptote) at a point in the integration interval. Improper integrals are evaluated as limits of proper definite integrals. *Example:* $\int_1^\infty \frac{1}{x^2} dx$ (infinite bound), $\int_0^1 \frac{1}{\sqrt{x}} dx$ (discontinuity at 0) ## Improper Integrals with Infinite Bounds (Type 1) Type 1 improper integrals have at least one infinite bound of integration, and the integrand is continuous on the entire interval. By definition: If $f$ is continuous on $[a, \infty)$: $$\int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx$$ If $f$ is continuous on $(-\infty, b]$: $$\int_{-\infty}^b f(x) dx = \lim_{a \to -\infty} \int_a^b f(x) dx$$ If $f$ is continuous on $(-\infty, \infty)$, split the integral at any finite point $c$: $$\int_{-\infty}^\infty f(x) dx = \int_{-\infty}^c f(x) dx + \int_c^\infty f(x) dx$$ The original integral converges **only if both separate integrals converge**. **Worked example:** Evaluate $\int_1^\infty \frac{1}{x^3} dx$ and state if it converges or diverges. 1. Rewrite the improper integral as a limit of proper integrals by definition: 2. $$\int_1^\infty \frac{1}{x^3} dx = \lim_{b \to \infty} \int_1^b x^{-3} dx$$ 3. Find the antiderivative of $x^{-3}$: 4. $$\int x^{-3} dx = -\frac{1}{2}x^{-2} + C = -\frac{1}{2x^2} + C$$ 5. Evaluate the definite integral from 1 to $b$: 6. $$\lim_{b \to \infty} \left[ -\frac{1}{2b^2} - \left(-\frac{1}{2(1)^2}\right) \right] = \lim_{b \to \infty} \left( \frac{1}{2} - \frac{1}{2b^2} \right)$$ 7. Evaluate the limit: as $b \to \infty$, $\frac{1}{2b^2} \to 0$, so the limit equals $\frac{1}{2}$. 8. The integral converges to $\frac{1}{2}$. > **tip** > > If you have an improper integral with infinite bounds on both ends, you must evaluate two separate limits. The Cauchy principal value ($\lim_{b \to \infty} \int_{-b}^b f(x)dx$) is not accepted as a valid convergence test on the AP exam. ## Improper Integrals with Discontinuous Integrands (Type 2) Type 2 improper integrals have finite integration bounds, but the integrand has an infinite discontinuity (vertical asymptote) at one or more points in the interval. We use the same limit-based approach, but approach the discontinuity instead of moving a bound to infinity. If $f$ is continuous on $(a, b]$ with an infinite discontinuity at $a$: $$\int_a^b f(x) dx = \lim_{c \to a^+} \int_c^b f(x) dx$$ If $f$ is continuous on $[a, b)$ with an infinite discontinuity at $b$: $$\int_a^b f(x) dx = \lim_{c \to b^-} \int_a^c f(x) dx$$ If the discontinuity is at an interior point $c \in (a,b)$, split the integral into two improper integrals at $c$; both must converge for the original integral to converge. **Worked example:** Evaluate $\int_0^8 \frac{1}{\sqrt[3]{x}} dx$ and state if it converges or diverges. 1. Identify the discontinuity: $\frac{1}{\sqrt[3]{x}}$ has an infinite discontinuity at $x=0$, the left endpoint of the interval. 2. Rewrite the improper integral as a right-hand limit by definition: 3. $$\int_0^8 \frac{1}{\sqrt[3]{x}} dx = \lim_{a \to 0^+} \int_a^8 x^{-1/3} dx$$ 4. Find the antiderivative of $x^{-1/3}$: 5. $$\int x^{-1/3} dx = \frac{3}{2}x^{2/3} + C$$ 6. Evaluate the definite integral from $a$ to 8: 7. $$\lim_{a \to 0^+} \left[ \frac{3}{2}(8^{2/3}) - \frac{3}{2}(a^{2/3}) \right] = \lim_{a \to 0^+} \left( 6 - \frac{3}{2}a^{2/3} \right)$$ 8. Evaluate the limit: as $a \to 0^+$, $a^{2/3} \to 0$, so the limit equals 6. 9. The integral converges to 6. > **tip** > > Always check for discontinuities in the integrand before integrating, even if the bounds are finite. AP exam questions often hide vertical asymptotes at interior points to test if you check the domain first. ## Convergence Testing: p-Test and Comparison Test On many AP exam questions, you only need to determine if an improper integral converges or diverges, not calculate its exact value. Two key tools for this are the p-test for power functions and the comparison test for general positive integrands. - **p-test for infinite bounds ($a>0$):** $\int_a^\infty \frac{1}{x^p} dx$ converges if $p>1$, diverges if $p \leq 1$. The function must decay fast enough as $x$ grows to have finite area. - **p-test for discontinuity at 0 ($a>0$):** $\int_0^a \frac{1}{x^p} dx$ converges if $p<1$, diverges if $p \geq 1$. The function cannot blow up too fast near $x=0$ to have finite area. The comparison test applies to positive functions: if $0 \leq f(x) \leq g(x)$ for all $x$ in the interval: - If $\int g(x) dx$ converges, then $\int f(x) dx$ also converges. - If $\int f(x) dx$ diverges, then $\int g(x) dx$ also diverges. **Worked example:** Without evaluating the integral, use the comparison test and p-test to determine if $\int_1^\infty \frac{3}{\sqrt{x}(x+1)} dx$ converges or diverges. 1. For all $x \geq 1$, $x+1 > x$, so we can bound the integrand above: 2. $$0 < \frac{3}{\sqrt{x}(x+1)} < \frac{3}{\sqrt{x} \cdot x} = \frac{3}{x^{3/2}}$$ 3. Recognize $\int_1^\infty \frac{3}{x^{3/2}} dx$ is a p-integral with $p = 3/2 > 1$. 4. By the p-test for infinite bounds, $\int_1^\infty \frac{3}{x^{3/2}} dx$ converges. 5. By the comparison test, since our integrand is positive and smaller than a convergent integrand, the original integral converges. > **tip** > > Memorize the two opposite p-test conditions: for infinite bounds, convergence requires $p>1$; for discontinuity at 0, convergence requires $p<1$. AP multiple-choice distractors almost always mix these two up. **Check your understanding** Test your understanding of the p-test: 1. Which of the following statements about $\int_0^1 \frac{e^{-x}}{x^{0.8}} dx$ is true? - A) The integral converges because $p=0.8 < 1$ by the p-test. - B) The integral converges because $p=0.8 > 1$ by the p-test. - C) The integral diverges because $p=0.8 < 1$ by the p-test. - D) The integral diverges because $p=0.8 > 1$ by the p-test. *Why:* Correct! The integral has an infinite discontinuity at $x=0$, so the p-test for bounded intervals tells us convergence when $p<1$. Since $0 < \frac{e^{-x}}{x^{0.8}} \leq \frac{1}{x^{0.8}}$, the comparison test confirms convergence. ## Mixed and Applied Improper Integral Problems **Worked example:** Let $f(x) = \frac{1}{(x-1)(x-3)}$. (a) Explain why $\int_0^4 f(x) dx$ is improper. (b) Split the integral into a sum of limits of proper integrals. (c) Determine if the integral converges or diverges. 1. (a) The integrand has infinite discontinuities (vertical asymptotes) at $x=1$ and $x=3$, both inside the integration interval $[0,4]$, so the integral is improper by definition. 2. (b) Split the integral at the two discontinuities, resulting in four separate limits: 3. $$\int_0^4 f(x) dx = \lim_{a \to 1^-} \int_0^a f(x) dx + \lim_{b \to 1^+} \int_b^2 f(x) dx + \lim_{c \to 3^-} \int_2^c f(x) dx + \lim_{d \to 3^+} \int_d^4 f(x) dx$$ 4. (c) Use partial fraction decomposition to get $f(x) = -\frac{1}{2(x-1)} + \frac{1}{2(x-3)}$. Evaluate the first limit: 5. $$\lim_{a \to 1^-} \int_0^a \left(-\frac{1}{2(x-1)} + \frac{1}{2(x-3)}\right) dx = \lim_{a \to 1^-} \left[ -\frac{1}{2}\ln|x-1| + \frac{1}{2}\ln|x-3| \right]_0^a$$ 6. As $a \to 1^-$, $\ln(1-a) \to -\infty$, so $-\frac{1}{2}\ln(1-a) \to +\infty$. The first limit diverges, so the entire integral diverges. **Worked example:** The total work required to move a 1000 kg spacecraft from Earth's surface to infinitely far away is given by: $W = \int_R^\infty \frac{GMm}{r^2} dr$, where $G = 6.67 \times 10^{-11}$, $M = 5.97 \times 10^{24}$, $m = 1000$, $R = 6.37 \times 10^6$. Calculate the total work in joules. 1. Rewrite the improper integral as a limit by definition: 2. $$W = GMm \lim_{b \to \infty} \int_R^b r^{-2} dr$$ 3. The antiderivative of $r^{-2}$ is $-r^{-1} = -\frac{1}{r}$. Evaluate the limit: 4. $$\lim_{b \to \infty} \left(-\frac{1}{b} + \frac{1}{R}\right) = 0 + \frac{1}{R} = \frac{1}{R}$$ 5. Substitute the given values: 6. $$W = \frac{GMm}{R} = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)}{6.37 \times 10^6} \approx 6.25 \times 10^{10}$$ 7. The total work (escape energy) is approximately $6.25 \times 10^{10}$ joules. ## Common pitfalls - **Wrong:** Evaluating $\int_{-\infty}^\infty x dx$ as $\lim_{b \to \infty} \int_{-b}^b x dx = 0$ and concluding convergence. - Why it fails: Confuses the Cauchy principal value with the formal AP definition of convergence for two-sided infinite improper integrals. - Correct: Always split the integral at a finite point and evaluate two separate limits. For this example, both limits diverge, so the original integral diverges. - **Wrong:** Applying the p-test to $\int_0^2 \frac{1}{x^3} dx$ and concluding convergence because $p=3>1$. - Why it fails: Mixes up the p-test conditions for infinite bounds vs. discontinuity at 0. - Correct: Recall the p-test for discontinuity at 0: $\int_0^a \frac{1}{x^p} dx$ converges only if $p<1$, so this integral diverges. - **Wrong:** Integrating $\int_{-1}^1 \frac{1}{x^2} dx$ directly with the Fundamental Theorem to get $-2$, ignoring the discontinuity at $x=0$. - Why it fails: Forgets to check for vertical asymptotes in the interior of the interval when bounds are finite. - Correct: Always check the integrand for domain restrictions and infinite discontinuities before integrating; split the integral at $x=0$ and evaluate two limits, both of which diverge here. - **Wrong:** Concluding that $\int_1^\infty \frac{1}{\sqrt{x} + 1} dx$ converges because $0 < \frac{1}{\sqrt{x} + 1} < \frac{1}{\sqrt{x}}$ and $\int_1^\infty \frac{1}{\sqrt{x}} dx$ diverges. - Why it fails: Misapplies the comparison test rules for convergence and divergence. - Correct: If your function is smaller than a divergent integral, you can conclude nothing. For this example, bound below: $\frac{1}{\sqrt{x}+1} > \frac{1}{2\sqrt{x}}$, so compare to the divergent integral $\int_1^\infty \frac{1}{2\sqrt{x}} dx$ to conclude divergence. - **Wrong:** Adding a divergent and convergent improper integral and concluding the whole integral converges because one part converges. - Why it fails: Incorrectly assumes convergence of one part offsets divergence of another. - Correct: If any of the split limits for an improper integral does not exist or is infinite, the entire improper integral diverges, regardless of other parts. ## Cheatsheet | Category | Formula/Rule | Notes | | --- | --- | --- | | Infinite Upper Bound | $\int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx$ | Converges iff limit is finite; $f$ continuous on $[a, \infty)$ | | Infinite Lower Bound | $\int_{-\infty}^b f(x) dx = \lim_{a \to -\infty} \int_a^b f(x) dx$ | Converges iff limit is finite; $f$ continuous on $(-\infty, b]$ | | Two Infinite Bounds | $\int_{-\infty}^\infty f(x) dx = \int_{-\infty}^c f(x) dx + \int_c^\infty f(x) dx$ | Converges only if both integrals converge; split at any finite $c$ | | Discontinuity at Left Endpoint | $\int_a^b f(x) dx = \lim_{c \to a^+} \int_c^b f(x) dx$ | $f$ continuous on $(a, b]$, infinite discontinuity at $a$ | | Discontinuity at Right Endpoint | $\int_a^b f(x) dx = \lim_{c \to b^-} \int_a^c f(x) dx$ | $f$ continuous on $[a, b)$, infinite discontinuity at $b$ | | p-Test (Infinite Bounds, $a>0$) | Converges: $p>1$; Diverges: $p \leq 1$ | For $\int_a^\infty \frac{1}{x^p} dx$ | | p-Test (Discontinuity at 0, $a>0$) | Converges: $p<1$; Diverges: $p \geq 1$ | For $\int_0^a \frac{1}{x^p} dx$ | | Comparison Test (Positive $f,g$) | If $0 \leq f(x) \leq g(x)$: (1) $\int g$ converges $\implies \int f$ converges; (2) $\int f$ diverges $\implies \int g$ diverges | Only applies to positive integrands; no conclusion if $f