Chapter 17 — Self-Assessment Quiz

Q: An improper integral over an infinite interval is defined as: The ordinary integral, with plugged into the antiderivative The limit of proper integrals as the upper bound runs to Always infinite Always zero

B. By definition . The integral converges only if that limit is finite. Section 17.1.

Q: Which value is correct: versus ? Both converge to The first converges to ; the second diverges Both diverge The first diverges; the second converges to

B. , while . A single notch in the exponent flips the verdict. Section 17.1.

Q: The p-integral converges if and only if: B) C) D)

B) . At infinity you need fast decay. The boundary case (the function ) diverges. Section 17.1.

Q: The p-integral converges if and only if: B) C) D)

C) . This is the mirror image of the rule at infinity: near zero you need only a mild blow-up. Again fails. Section 17.2.

Q: Evaluate . Diverges B) C) D)

C) . The integrand blows up at , but , so the singularity is integrable: . Section 17.2.

Q: A student writes . The correct verdict is: B) by symmetry C) Diverges D)

C) Diverges. The integrand explodes at the interior point , so FTC does not apply. Splitting at shows each half is a p-integral at zero with , which diverges. (A positive integrand could never yield anyway.) Section 17.2.

Q: To show converges by direct comparison, the cleanest move is: Compare with , which diverges Compare with , which converges, since Evaluate the antiderivative exactly Conclude divergence because

B. For , , and converges (). A smaller positive function cannot enclose more area than a convergent one. Section 17.3.

Q: Using limit comparison, should be compared with because: The numerator has degree Keeping leading terms gives , and the ratio limit is finite and positive always converges The integrand is even

B. Leading-term behavior is ; , finite and positive, so the integral shares the fate of — it converges. Section 17.3.

Q: The Gamma function satisfies and . Therefore, for a positive integer , and for the famous half-integer value: and and and and

B. The recursion plus gives ; the substitution links to the Gaussian integral, yielding . Section 17.4.

Q: The Gaussian integral is remarkable because: Its antiderivative is , a simple polynomial has no elementary antiderivative, yet the definite integral has the exact value (proved in Chapter 32) It diverges It equals before normalization

B. By Liouville's theorem has no elementary antiderivative, so the FTC bar is useless; the value is obtained by squaring and switching to polar coordinates (Chapter 32). This is the source of the normalizing the normal density. Section 17.5.

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Chapter 17 — Self-Assessment Quiz

10 questions, ~20 minutes. Aim for 8/10. Each answer cites the section to revisit if you miss it.

1. An improper integral over an infinite interval is defined as: - A) The ordinary integral, with $\infty$ plugged into the antiderivative - B) The limit of proper integrals as the upper bound runs to $\infty$ - C) Always infinite - D) Always zero

Answer

B. By definition $\int_a^\infty f = \lim_{t\to\infty}\int_a^t f$. The integral converges only if that limit is finite. Section 17.1.

2. Which value is correct: $\displaystyle\int_1^\infty \frac{1}{x^2}\,dx$ versus $\displaystyle\int_1^\infty \frac{1}{x}\,dx$? - A) Both converge to $1$ - B) The first converges to $1$; the second diverges - C) Both diverge - D) The first diverges; the second converges to $1$

Answer

B. $\int_1^t x^{-2}\,dx = 1 - 1/t \to 1$, while $\int_1^t x^{-1}\,dx = \ln t \to \infty$. A single notch in the exponent flips the verdict. Section 17.1.

3. The p-integral $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx$ converges if and only if: - A) $p > 0$ B) $p > 1$ C) $p < 1$ D) $p \ge 1$

Answer

B) $p > 1$. At infinity you need fast decay. The boundary case $p=1$ (the function $1/x$) diverges. Section 17.1.

4. The p-integral $\displaystyle\int_0^1 \frac{1}{x^p}\,dx$ converges if and only if: - A) $p > 1$ B) $p \ge 1$ C) $p < 1$ D) $p = 1$

Answer

C) $p < 1$. This is the mirror image of the rule at infinity: near zero you need only a mild blow-up. Again $p=1$ fails. Section 17.2.

5. Evaluate $\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$. - A) Diverges B) $1$ C) $2$ D) $1/2$

Answer

C) $2$. The integrand blows up at $0$, but $p=1/2 < 1$, so the singularity is integrable: $[2\sqrt{x}]_0^1 = 2$. Section 17.2.

6. A student writes $\displaystyle\int_{-1}^1 \frac{1}{x^2}\,dx = \big[-\tfrac1x\big]_{-1}^1 = -2$. The correct verdict is: - A) $-2$ B) $0$ by symmetry C) Diverges D) $2$

Answer

C) Diverges. The integrand explodes at the interior point $x=0$, so FTC does not apply. Splitting at $0$ shows each half is a p-integral at zero with $p=2 \ge 1$, which diverges. (A positive integrand could never yield $-2$ anyway.) Section 17.2.

7. To show $\displaystyle\int_1^\infty \frac{1}{1+x^3}\,dx$ converges by direct comparison, the cleanest move is: - A) Compare with $1/x$, which diverges - B) Compare with $1/x^3$, which converges, since $\frac{1}{1+x^3} < \frac{1}{x^3}$ - C) Evaluate the antiderivative exactly - D) Conclude divergence because $1+x^3 > x^3$

Answer

B. For $x\ge 1$, $\frac{1}{1+x^3} < \frac{1}{x^3}$, and $\int_1^\infty x^{-3}\,dx$ converges ($p=3>1$). A smaller positive function cannot enclose more area than a convergent one. Section 17.3.

8. Using limit comparison, $\displaystyle\int_1^\infty \frac{2x+1}{x^3+5}\,dx$ should be compared with $g(x)=1/x^2$ because: - A) The numerator has degree $1$ - B) Keeping leading terms gives $\frac{2x}{x^3}=\frac{2}{x^2}$, and the ratio limit is finite and positive - C) $1/x^2$ always converges - D) The integrand is even

Answer

B. Leading-term behavior is $2/x^2$; $\lim_{x\to\infty}\frac{(2x+1)/(x^3+5)}{1/x^2}=2$, finite and positive, so the integral shares the fate of $\int_1^\infty x^{-2}\,dx$ — it converges. Section 17.3.

9. The Gamma function satisfies $\Gamma(s+1)=s\,\Gamma(s)$ and $\Gamma(1)=1$. Therefore, for a positive integer $n$, and for the famous half-integer value: - A) $\Gamma(n)=n!$ and $\Gamma(1/2)=\pi$ - B) $\Gamma(n)=(n-1)!$ and $\Gamma(1/2)=\sqrt{\pi}$ - C) $\Gamma(n)=n!$ and $\Gamma(1/2)=\sqrt{\pi}$ - D) $\Gamma(n)=(n-1)!$ and $\Gamma(1/2)=\pi$

Answer

B. The recursion plus $\Gamma(1)=1$ gives $\Gamma(n)=(n-1)!$; the substitution $x=u^2$ links $\Gamma(1/2)$ to the Gaussian integral, yielding $\sqrt{\pi}$. Section 17.4.

10. The Gaussian integral $\displaystyle\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$ is remarkable because: - A) Its antiderivative is $\operatorname{erf}$, a simple polynomial - B) $e^{-x^2}$ has no elementary antiderivative, yet the definite integral has the exact value $\sqrt{\pi}$ (proved in Chapter 32) - C) It diverges - D) It equals $\sqrt{2\pi}$ before normalization

Answer

B. By Liouville's theorem $e^{-x^2}$ has no elementary antiderivative, so the FTC bar is useless; the value $\sqrt{\pi}$ is obtained by squaring and switching to polar coordinates (Chapter 32). This is the source of the $\sqrt{2\pi}$ normalizing the normal density. Section 17.5.

Scoring

9–10: Excellent. You command both the evaluation machinery and the convergence tests.
7–8: Solid. Review any missed item's cited section.
5–6: Re-read Sections 17.1 and 17.2 (the two p-integral rules — the most common stumbling block) and Section 17.3 (comparison tests).
Below 5: Slow down and rework the chapter's worked examples by hand. This material is the critical foundation for probability, statistics (Section 17.5), and the integral test for series in Chapter 22.