AP Calculus BC 6.13 Evaluating Improper Integrals -MCQs - Exam Style Questions
No-CalcQuestion
Compute the improper integral \[ \int_{1}^{\infty} x\,e^{-x}\,dx . \]
(A) \(0\)
(B) \(\tfrac{1}{e}\)
(C) \(\tfrac{2}{e}\)
(D) nonexistent
▶️ Answer/Explanation
Detailed solution
Integration by parts on \([1,\infty)\). Let \(u=x\), \(dv=e^{-x}dx\Rightarrow v=-e^{-x}\).
\(\displaystyle \int x e^{-x}\,dx=-x e^{-x}-e^{-x}+C.\)
\[ \int_{1}^{\infty} x e^{-x}\,dx =\lim_{b\to\infty}\big[-(x+1)e^{-x}\big]_{1}^{b} =0-[-2e^{-1}] =\tfrac{2}{e}. \] ✅ Answer: (C)
No-CalcQuestion
Evaluate the improper integral \(\displaystyle \int_{3}^{5}\frac{x}{x^{2}-9}\,dx\).
(A) \(\dfrac{1}{2}\ln\!\left(\dfrac{5}{3}\right)\)
(B) \(\ln\!\left(\dfrac{5}{3}\right)\)
(C) \(\dfrac{1}{2}\ln(16)\)
(D) divergent
▶️ Answer/Explanation
Correct Answer: D
The integrand is unbounded at \(x=3\) since \(x^{2}-9=(x-3)(x+3)\). Treat as an improper integral: \[ \int_{3}^{5}\frac{x}{x^{2}-9}\,dx =\lim_{a\to 3^{+}}\int_{a}^{5}\frac{x}{x^{2}-9}\,dx =\lim_{a\to 3^{+}}\frac{1}{2}\ln|x^{2}-9|\Big|_{a}^{5}. \]
At \(x=5\): \(\tfrac{1}{2}\ln(16)\) is finite. As \(a\to 3^{+}\): \(\ln(a^{2}-9)\to -\infty\). The difference \(\tfrac{1}{2}\ln(16)-\left(-\infty\right)=+\infty\).
Therefore, the integral diverges.