## H1011 Quizzes

Quiz 11: One variable integration
Question 1 Questions
Evaluate ${\int }_{0}^{1}\left(6{x}^{2}-4x+3\right)\phantom{\rule{1em}{0ex}}dx$.
 a) 4 b) -3 c) 3 d) 5

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
${\int }_{0}^{1}\left(6{x}^{2}-4x+3\right)\phantom{\rule{1em}{0ex}}dx={\left[2{x}^{3}-2{x}^{2}+3x\right]}_{0}^{1}=2-2+3=3.$
Choice (d) is incorrect
Evaluate ${\int }_{0}^{\frac{\pi }{2}}\left(sinx+cosx\right)\phantom{\rule{1em}{0ex}}dx$.
 a) 2 b) 0 c) $\frac{\pi }{2}$ d) -2

Choice (a) is correct!
$\begin{array}{rcll}{\int }_{0}^{\frac{\pi }{2}}\left(sinx+cosx\right)\phantom{\rule{1em}{0ex}}dx& =& {\left[-cosx+sinx\right]}_{0}^{\frac{\pi }{2}}& \text{}\\ & =& -cos\frac{\pi }{2}+sin\frac{\pi }{2}-\left(-cos0+sin0\right)& \text{}\\ & =& 1+1=2.& \text{}\end{array}$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Evaluate ${\int }_{0}^{\frac{\pi }{4}}sin2x\phantom{\rule{1em}{0ex}}dx$.
 a) 1 b) $\frac{1}{2}$ c) $-\frac{1}{2}$ d) -1

Choice (a) is incorrect
Choice (b) is correct!
$\begin{array}{rcll}{\int }_{0}^{\frac{\pi }{4}}sin2x\phantom{\rule{1em}{0ex}}dx& =& {\left[-\frac{1}{2}cos2x\right]}_{0}^{\frac{\pi }{4}}& \text{}\\ & =& -\frac{1}{2}cos\frac{\pi }{2}+\frac{1}{2}cos0& \text{}\\ & =& \frac{1}{2}.& \text{}\end{array}$
Choice (c) is incorrect
Choice (d) is incorrect
Evaluate ${\int }_{0}^{\sqrt{3}}x{\left(1+{x}^{2}\right)}^{\frac{3}{2}}\phantom{\rule{1em}{0ex}}dx$.
 a) $\frac{32}{5}$ b) $\frac{31}{5}$ c) $\frac{3}{2}$ d) $\frac{9}{2}$

Choice (a) is incorrect
Choice (b) is correct!
$\begin{array}{rcll}{\int }_{0}^{\sqrt{3}}x{\left(1+{x}^{2}\right)}^{\frac{3}{2}}\phantom{\rule{1em}{0ex}}dx& =& {\left[\frac{1}{5}{\left(1+{x}^{2}\right)}^{\frac{5}{2}}\right]}_{0}^{\sqrt{3}}& \text{}\\ & =& \frac{1}{5}{\left(4\right)}^{\frac{5}{2}}-\frac{1}{5}& \text{}\\ & =& \frac{32}{5}-\frac{1}{5}=\frac{31}{5}.& \text{}\end{array}$
Choice (c) is incorrect
Choice (d) is incorrect
Evaluate ${\int }_{0}^{2}3{x}^{2}{e}^{{x}^{3}}\phantom{\rule{1em}{0ex}}dx$.
 a) ${e}^{8}-1$ b) $8{e}^{8}$ c) $-{e}^{8}$ d) $12{e}^{8}-12$

Choice (a) is correct!
${\int }_{0}^{2}3{x}^{2}{e}^{{x}^{3}}\phantom{\rule{1em}{0ex}}dx={\left[{e}^{{x}^{3}}\right]}_{0}^{2}={e}^{8}-{e}^{0}={e}^{8}-1.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Let $A$ be the area under the curve $f\left(x\right)=16-{x}^{2}$ on the interval [0,4]. Dividing the interval into 4 subintervals and defining ${x}_{i}$ to be the midpoint of the $i$th interval (the area of the rectangle is thus $f\left({x}_{i}\right)$ times the length of the interval), the best estimate for $A$ is
 a) $A={\sum }_{i=1}^{4}\left(16-{x}_{i}^{2}\right)×\frac{1}{2}$, where ${x}_{i}=\frac{i-1}{2}$ b) $A={\sum }_{i=1}^{4}\left(16-{x}_{i}^{2}\right)×\frac{1}{2}$, where ${x}_{i}=\frac{i-1}{2}$ c) $A={\sum }_{i=1}^{4}\left(16-{x}_{i}^{2}\right)$, where ${x}_{i}=\frac{2i-1}{2}$ d) $A={\sum }_{i=0}^{3}\left(16-{x}_{i}^{2}\right)$, where ${x}_{i}=\frac{2i-1}{2}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The intervals are of length 1, and the area is calculated at $\frac{1}{2}$, $\frac{3}{2}$, $\frac{5}{2}$ and $\frac{7}{2}$.
Choice (d) is incorrect
Which of the following sums is the best estimate for $A$, the area under the curve $f\left(x\right)={x}^{3}+2$ on the interval [-1,2], divided into 6 subintervals and choosing ${x}_{i}$ as the right-endpoint of the $i$th interval ?
 a) $A={\sum }_{i=0}^{6}\left({x}_{i}^{3}+2\right)$, where ${x}_{i}=\frac{i-3}{2}$ b) $A={\sum }_{i=0}^{6}\left({x}_{i}^{3}+2\right)×\frac{1}{2}$, where ${x}_{i}=\frac{2i-1}{2}$ c) $A={\sum }_{i=1}^{6}\left({x}_{i}^{3}+2\right)$, where ${x}_{i}=\frac{2i-3}{2}$ d) $A={\sum }_{i=1}^{6}\left({x}_{i}^{3}+2\right)×\frac{1}{2}$, where ${x}_{i}=\frac{i-2}{2}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The intervals are of length $\frac{1}{2}$, and the area is calculated at $-\frac{1}{2}$, $0$, $\frac{1}{2}$, $1$, $\frac{3}{2}$, $2$
Find the indefinite integral $\int 2{t}^{2}{\left(1+{t}^{3}\right)}^{4}\phantom{\rule{1em}{0ex}}dt.$
 a) $\frac{1}{5}{\left(1+{t}^{3}\right)}^{5}+C$ b) $\frac{2}{5}{\left(1+{t}^{3}\right)}^{5}+C$ c) $\frac{2}{3}{t}^{3}{\left(1+{t}^{3}\right)}^{5}+C$ d) $\frac{2}{15}{\left(1+{t}^{3}\right)}^{5}+C$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Let $u=1+{t}^{3},du=3{t}^{2}\phantom{\rule{1em}{0ex}}dt$. $\begin{array}{rcll}\int 2{t}^{2}{\left(1+{t}^{3}\right)}^{4}dt& =& \frac{2}{3}\int {u}^{4}\phantom{\rule{1em}{0ex}}du& \text{}\\ & =& \frac{2}{15}{u}^{5}+C& \text{}\\ & =& \frac{2}{15}{\left(1+{t}^{3}\right)}^{5}+C.& \text{}\end{array}$
Find the indefinite integral $\int {sin}^{3}xcosx\phantom{\rule{1em}{0ex}}dx.$
 a) ${sin}^{4}x+C$ b) $\frac{1}{4}{sin}^{4}x+C$ c) $3{sin}^{2}x+C$ d) $-\frac{1}{4}{sin}^{4}x+C$

Choice (a) is incorrect
Choice (b) is correct!
Let $u=sinx$, $du=cosx\phantom{\rule{1em}{0ex}}dx.$ $\begin{array}{rcll}\int {sin}^{3}xcosx\phantom{\rule{1em}{0ex}}dx& =& \int {u}^{3}\phantom{\rule{1em}{0ex}}du& \text{}\\ & =& \frac{1}{4}{u}^{4}+C& \text{}\\ & =& \frac{1}{4}{sin}^{4}x+C.& \text{}\end{array}$
Choice (c) is incorrect
Choice (d) is incorrect
Find the indefinite integral
$\int \frac{2t-{e}^{-t}+4}{{\left({t}^{2}+4t+{e}^{-t}+1\right)}^{2}}\phantom{\rule{1em}{0ex}}dt.$
 a) $\frac{1}{{t}^{2}+4t+{e}^{-t}+1}+C$ b) $\frac{-1}{3{\left({t}^{2}+4t+{e}^{-t}+1\right)}^{3}}+C$ c) $\frac{-1}{{t}^{2}+4t+{e}^{-t}+1}+C$ d) $\frac{1}{3{\left({t}^{2}+4t+{e}^{-t}+1\right)}^{3}}+C$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Let $u={t}^{2}+4t+{e}^{-t}+1$, $du=\left(2t+4-{e}^{-t}\right)\phantom{\rule{1em}{0ex}}dt.$ $\begin{array}{rcll}\int \frac{2t-{e}^{-t}+4}{{\left({t}^{2}+4t+{e}^{-t}+1\right)}^{2}}\phantom{\rule{1em}{0ex}}dt& =& \int \frac{1}{{u}^{2}}\phantom{\rule{1em}{0ex}}du& \text{}\\ & =& -{u}^{-1}+C& \text{}\\ & =& \frac{-1}{{t}^{2}+4t+{e}^{-t}+1}+C.& \text{}\end{array}$
Choice (d) is incorrect