Question
Choose the Riemann Sum whose limit is the integral $\int_{-1}^3 \ln \left(x^2+2\right) d x$.
(A) $\lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\ln \left(\left(\frac{4 k}{n}-1\right)^2+2\right) \cdot\left(\frac{4}{n}\right)\right)$
(B) $\lim _{n \rightarrow \infty} \sum_{k=1}^{k=\bar{n}}\left(\ln \left(\left(\frac{4 k}{n}-1\right)^2+2\right) \cdot\left(\frac{1}{n}\right)\right)$
(C) $\lim _{n \rightarrow \infty} \sum_{k=1}^{k=\bar{n}}\left(\ln \left(\left(\frac{k}{n}-1\right)^2+2\right) \cdot\left(\frac{1}{n}\right)\right)$
(D) $\lim _{n \rightarrow \infty} \sum_{k=1}^{n=1}\left(\ln \left(\left(\frac{k}{n}-1\right)^2+2\right) \cdot\left(\frac{4}{n}\right)\right)$
▶️Answer/Explanation
Ans:A
From the integral we get $a=-1, b=3$, so $\Delta x=\frac{3-(-1)}{n}=\frac{4}{n}$ and $x_k=a+k \cdot \Delta x=-1+\frac{4 k}{n}=\frac{4 k}{n}-1$. Replace $x$ with $x_k$ and replace $d x$ with $\Delta x$ in the integrand to get the general term in the summation.
Question
A particle moves on a line according to the law $s=f(t)$ so that its velocity is $v=k s$, where $k$ is a nonzero constant. The acceleration of the particle is
(A) $k^2 v$
(B) $k^2 s$
(C) $k$
(D) 0
▶️Answer/Explanation
Ans:B
Since $v=k s=\frac{d s}{d t}$ then $a=\frac{d^2 s}{d t^2}=k \frac{d^2 s}{d t^2}=k \frac{d s}{d t}=k v=k^2 s$.
Question
A cup of coffee placed on a table cools at a rate of $\frac{d H}{d t}=-0.05(H-70)^{\circ} \mathrm{F}$ per minute, where $H$ represents the temperature of the coffee and $t$ is time in minutes. If the coffee was at $120^{\circ} \mathrm{F}$ initially, what will its temperature be, to the nearest degree, 10 minutes later?
(A) $73^{\circ} \mathrm{F}$
(B) $95^{\circ} \mathrm{F}$
(C) $100^{\circ} \mathrm{F}$
(D) $105^{\circ} \mathrm{F}$
▶️Answer/Explanation
Ans:C
$\begin{aligned} \frac{d H}{H-70}=-0.05 d t . \quad \ln |H-70| & =-0.05 t+C \\ H-70 & =c e^{-0.05 t} \\ H(x) & =70+c e^{-0.05 t}\end{aligned}$
The initial condition $H(0)=120$ shows $c=50$. Evaluate $H(10)$.
Question
If $f(x)=\int_0^{x^2+2} \sqrt{1+\cos t} d t$, then $f(x)=$
(A) $2 x \sqrt{1+\cos \left(x^2+2\right)}$
(B) $2 x \sqrt{1-\sin x}$
(C) $\sqrt{1+\cos \left(x^2+2\right)}$
(D) $\sqrt{\left[1+\cos \left(x^2+2\right)\right] \cdot 2 x}$
▶️Answer/Explanation
Ans:A
Let $u=x^2+2$. Then
$
\frac{d}{d u} \int_0^u \sqrt{1+\cos t} d t=\sqrt{1+\cos u}
$
and $
\frac{d}{d x} \int_0^u \sqrt{1+\cos t} d t=\sqrt{1+\cos u} \frac{d u}{d x}=\sqrt{1+\cos \left(x^2+2\right)} \cdot(2 x)
$
Question
$\lim _{x \rightarrow \infty} \frac{3+x-2 x^2}{4 x^2+9}$ is
(A) $-\frac{1}{2}$
(B) $\frac{1}{2}$
(C) 1
(D) $\frac{3}{4}$
▶️Answer/Explanation
Ans:A
Since the degrees of the numerator and the denominator are the same, the limit as $x \rightarrow \infty$ is the ratio of the coefficients of the terms of highest degree: $\frac{-2}{4}$.