## MATH1111 Quizzes

The Definite Integral Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 21 and corresponds to Section 5.2 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are no Wiley web quizzes for this topic.

If you need some help on sigma notation then go to http://quiz.econ.usyd.edu.au/mathquiz/sigma/index.php for quizzes and web resources.

Questions 6 and 7 at http://www.maths.usyd.edu.au/u/UG/JM/MATH1011/Quizzes/quiz10.html could be useful.

There is an excellent animation of the general Riemann sum at http://archives.math.utk.edu/visual.calculus/4/riemann_sums.4/index.html.

Which of the following is the left hand sum, with $n=4\phantom{\rule{0.3em}{0ex}},$ for ${\int }_{1}^{5}\left(1+{x}^{2}\right)dx\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) 34 b) 58 c) 60 d) 44.33

Choice (a) is correct!
Left hand sum ${\sum }_{i=0}^{3}f\left({x}_{i}\right)\Delta x={\sum }_{i=0}^{3}\left(1+{x}_{i}\right)\Delta x$ where $\Delta x=\frac{5-1}{4}=1\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{x}_{0}^{}=1$ and ${x}_{4}^{}=5\phantom{\rule{0.3em}{0ex}}.$

So the left hand sum $=f\left(1\right)×1+f\left(2\right)×1+f\left(3\right)×1+f\left(4\right)×1=2+5+10+17=34$
Choice (b) is incorrect
Try again, this is the right hand sum $=f\left(2\right)+f\left(3\right)+f\left(4\right)+f\left(5\right)=5+10+17+26=58$
Choice (c) is incorrect
Try again, this is $=f\left(1\right)+f\left(2\right)+f\left(3\right)+f\left(4\right)+f\left(5\right)=2+5+10+17+26=60$ which has one term too many.
Choice (d) is incorrect
Try again, this is the exact answer for ${\int }_{1}^{5}\left(1+{x}^{2}\right)dx\phantom{\rule{0.3em}{0ex}}.$
We wish to find the right hand sum of ${\int }_{-3}^{2}{t}^{3}\phantom{\rule{0.3em}{0ex}}dt$ with $n=10\phantom{\rule{0.3em}{0ex}}.$
Which of the values of ${t}_{i}$ below need to be substituted into ${\sum }_{i=1}^{10}{\left({t}_{i}\right)}^{3}\Delta t\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) ${t}_{0}^{}=-3\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{1}^{}=-2.5\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{2}^{}=-2\phantom{\rule{0.3em}{0ex}},\dots \phantom{\rule{0.3em}{0ex}},{t}_{9}^{}=2\phantom{\rule{0.3em}{0ex}}.$ b) ${t}_{0}^{}=-3\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{1}^{}=-2\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{1}^{}=-1\phantom{\rule{0.3em}{0ex}},\dots \phantom{\rule{0.3em}{0ex}},{t}_{9}^{}=1\phantom{\rule{0.3em}{0ex}}.$ c) ${t}_{1}^{}=-2\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{2}^{}=-1\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{3}^{}=0\phantom{\rule{0.3em}{0ex}},\dots \phantom{\rule{0.3em}{0ex}},{t}_{10}^{}=2\phantom{\rule{0.3em}{0ex}}.$ d) ${t}_{1}^{}=-2.5\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{1}^{}=-2\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{t}_{2}^{}=-1.5\phantom{\rule{0.3em}{0ex}},\dots \phantom{\rule{0.3em}{0ex}},{t}_{10}^{}=2\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, this is what you would require for a left hand sum.
Choice (b) is incorrect
Try again, you need $\Delta t=0.5\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, you need $\Delta t=0.5\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is correct!
Since The right hand sum gives this approximation

we need to start the sum at ${t}_{1}^{}=-2.5$ and increase each term by $\Delta t=\frac{2-\left(-3\right)}{10}=0.5$ to end at ${t}_{10}^{}=2\phantom{\rule{0.3em}{0ex}}.$
Which of the following is the right hand sum, with $n=6\phantom{\rule{0.3em}{0ex}},$ for ${\int }_{-1}^{2}\left(3{x}^{2}+2x+1\right)\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) 11.625 b) 19.125 c) 20.125 d) 13

Choice (a) is incorrect
Try again, this is the left hand sum.
Choice (b) is correct!
Right hand sum ${\sum }_{i=1}^{6}f\left({x}_{i}\right)\Delta x={\sum }_{i=1}^{6}\left(3{x}_{i}^{2}+2{x}_{i}+1\right)\Delta x$ where $\Delta x=\frac{2+1}{6}=0.5\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{x}_{0}^{}=-1$ and ${x}_{6}^{}=2\phantom{\rule{0.3em}{0ex}}.$

So the right hand sum
$=0.5\left(f\left(-0.5\right)+f\left(0\right)+f\left(0.5\right)+f\left(1\right)+f\left(1.5\right)+f\left(2\right)\right)$ $=0.5\left(0.75+1+2.75+6+10.75+17\right)=19.125$
Choice (c) is incorrect
Try again, there are too many terms
Choice (d) is incorrect
Try again, this is the exact value of ${\int }_{-1}^{2}\left(3{x}^{2}+2x+1\right)\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}.$
Which of the following would be the appropriate integral to evaluate if you were asked to find the area between the curve $y=16-{x}^{4}$ and the $x$-axis? Exactly one option must be correct)
 a) ${\int }_{-2}^{2}\left(16-{x}^{4}\right)\phantom{\rule{0.3em}{0ex}}dx$ b) ${\int }_{-4}^{4}\left(16-{x}^{4}\right)\phantom{\rule{0.3em}{0ex}}dx$ c) ${\int }_{2}^{-2}\left(16-{x}^{4}\right)\phantom{\rule{0.3em}{0ex}}dx$ d) ${\int }_{0}^{16}\left(16-{x}^{4}\right)\phantom{\rule{0.3em}{0ex}}dx$

Choice (a) is correct!
The curve $y=16-{x}^{4}=\left({x}^{2}-4\right)\left({x}^{2}+4\right)=\left(x-2\right)\left(x+2\right)\left({x}^{2}+4\right)$ cuts the $x$-axis at $x=-2$ and $x=2$ so this is the most appropriate integral.
Choice (b) is incorrect
Try again, you need to factorize $16-{x}^{4}$ to determine where it cuts the $x$-axis.
Choice (c) is incorrect
Try again, this will give you a negative value for the area.
Choice (d) is incorrect
Try again, you need to factorize $16-{x}^{4}$ to determine where it cuts the $x$-axis.