What is the area bounded by the graphs of
$f\left(x\right)=sinx$ and
$g\left(x\right)=cosx$ and the
lines $x=0$ and
$x=\frac{\pi}{2}$ ? Exactly one option
must be correct)

*Choice (a) is correct!*

The area is given by $2{\int}_{0}^{\pi \u22154}\phantom{\rule{1em}{0ex}}\left(cosx-sinx\right)\phantom{\rule{1em}{0ex}}dx$
and this equals $2\left(\sqrt{2}-1\right).$

*Choice (b) is incorrect*

Have you
sketched the graphs and found their point of intersection? Try to use symmetry to make your work
easier.

*Choice (c) is incorrect*

Have you
sketched the graphs and found their point of intersection? Try to use symmetry to make your
work easier.

*Choice (d) is incorrect*

Have you sketched the graphs and found their point of intersection? Try to use
symmetry to make your work easier.

*Choice (e) is incorrect*

What is the area bounded by the graphs of
$f\left(x\right)=\sqrt{x-1}$,
$g\left(x\right)=\frac{2}{x}$, the
$x$- axis and the
line $x=3$ ? Enter
your answer correct to 2 decimal places. Do not enter any units.

*Correct!*

Well done. The required
area is ${\int}_{1}^{2}\phantom{\rule{1em}{0ex}}\sqrt{x-1}\phantom{\rule{1em}{0ex}}dx+{\int}_{2}^{3}\phantom{\rule{1em}{0ex}}\frac{2}{x}\phantom{\rule{1em}{0ex}}dx$ and
this equals $\frac{2}{3}+2ln\frac{3}{2}\approx 1.48$.

*Incorrect.*

*Please try again.*

Sketch the curves and find their point of intersection. Now express the required area
as the sum of two separate integrals.

Which option below is an integral which equals the volume of
the solid of revolution formed when the area between the curves
$y=\sqrt{4x}$ and
$y=2{x}^{3}$ and between
the lines $x=0$ and
$x=1$ is rotated about the
line $x=3$ ? Exactly one
option must be correct)

*Choice (a) is incorrect*

Which method are you thinking about - shells or discs? Check that you have the
right expression to integrate and the right endpoints on the integral sign.

*Choice (b) is correct!*

This is the
integral which gives the required volume when the method of shells is used. The radius of the
shell is $3-x$ and the
height $\sqrt{4x}-2{x}^{3}$. Values of
$x$ range from 0 to 1.

*Choice (c) is incorrect*

Check the limits on the
integral sign.

*Choice (d) is incorrect*

Which is greater
on the interval concerned, $\sqrt{4x}$
or $2{x}^{3}$ ?

The method of shells is used to obtain the volume
$V$
of the solid of revolution formed when the area between the curve
$y={x}^{2}$ and the
$x$-axis,
from $x=0$ to
$x=1$, is rotated
about the line $y=-2$.
This method gives us
$$V=2\pi {\int}_{0}^{1}\phantom{\rule{1em}{0ex}}\left(2+y\right)\left(1-\sqrt{y}\right)\phantom{\rule{1em}{0ex}}dy.$$
Which of the integrals below is the one which calculates the same
volume by the method of discs? Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

*Choice (d) is incorrect*

*Choice (e) is incorrect*

Use a substitution to find $\int \phantom{\rule{1em}{0ex}}\frac{t}{\sqrt{4-{t}^{4}}}\phantom{\rule{1em}{0ex}}dt.$ Exactly
one option must be correct)

*Choice (a) is incorrect*

You can check if your answer is right by differentiating it. Do this and you will see why this option
is not correct.

*Choice (b) is incorrect*

You
can check if your answer is right by differentiating it. Do this and you will see why this option is not
correct.

*Choice (c) is incorrect*

It is not
valid to simplify $\sqrt{4-{t}^{4}}$
to $\sqrt{4}-\sqrt{{t}^{4}}$.

*Choice (d) is incorrect*

You can check
if your answer is right by differentiating it. Do this and you will see why this option is not
correct.

*Choice (e) is correct!*

Useful
substitutions are $u={t}^{2}$
or $u=\frac{{t}^{2}}{2}$.

Use a substitution to find $\int \phantom{\rule{1em}{0ex}}\sqrt{tanx}\phantom{\rule{1em}{0ex}}{sec}^{4}x\phantom{\rule{1em}{0ex}}dx.$
Hint: try a substitution that gets rid of the square root sign. Exactly one option must be
correct)

*Choice (a) is incorrect*

Differentiate your answer to see why this is incorrect. Try the substitution
$u=\sqrt{tanx}$.

*Choice (b) is incorrect*

Differentiate your answer to see why this is incorrect. Try the substitution
$u=\sqrt{tanx}$.

*Choice (c) is correct!*

The substitution
$u=\sqrt{tanx}$, together with the
identity $1+{tan}^{2}x={sec}^{2}x$ are useful
in this problem.

*Choice (d) is incorrect*

Differentiate your answer to see why this is incorrect. Try the substitution
$u=\sqrt{tanx}$.

*Choice (e) is incorrect*

Use the shell method to find the volume of the solid formed when the area enclosed by the
curves $y={x}^{2}-4x+3$ and
$y=-{x}^{2}+2x+3$ is rotated about
the $y$-axis. Enter
your answer correct to two decimal places. (Do not enter any units.)

*Correct!*

Well done! Your integral
should be $2\pi {\int}_{0}^{3}\phantom{\rule{1em}{0ex}}x\left(-2{x}^{2}+6x\right)\phantom{\rule{1em}{0ex}}dx=27\pi ,$ which is
84.82 correct to two decimal places.

*Incorrect.*

*Please try again.*

To use the shell method, you need the radius (in this case, just
$x$) and the height of the
shell (in this case, $\left(-{x}^{2}+2x+3\right)-\left({x}^{2}-4x+3\right)$).
Now put this information together to obtain the volume of the shell, and then
integrate.

Use the disc method to find the volume of the solid formed when the area enclosed by the
curve$y=sinx$ and
the $x$-axis,
between $x=0$ and
$x=\pi $, is rotated
about the line $y=2$.
Enter your answer correct to two decimal places. (Do not enter any
units.)

*Correct!*

*Incorrect.*

*Please try again.*

The disc method requires us to set up an integral describing
a disc with a hole, formed by rotating the area about the line
$y=2$.
The outer radius of the disc is 2 and the inner radius is
$2-sinx$.

Find the volume of the solid formed when the area enclosed by the
curves$y=e{x}^{2}$ and
$y={e}^{x}$, and the line
$x=0$, is rotated about the
$y$-axis. You may use the fact
that $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx={e}^{x}\left(x-1\right)+C.$ Exactly one option must
be correct)

*Choice (a) is correct!*

The volume
is given by the integral $2\pi {\int}_{0}^{1}\phantom{\rule{1em}{0ex}}x\left({e}^{x}-e{x}^{2}\right)\phantom{\rule{1em}{0ex}}dx.$
When evaluated using the hint in the question, this equals
$2\pi (1-\frac{e}{4}$.

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is incorrect*

*Choice (e) is incorrect*

What is the volume of the solid obtained when the area between the
$x$-axis and
the curve $y={x}^{3}+1$,
from $x=-1$ to
$x=1$, is rotated about the
line $y=-1$ ? Exactly one
option must be correct)
Hint: the disc method is easier here than the shell method. The inner radius
is 1. What is the outer radius?

*Choice (a) is incorrect*

Hint: the disc method is easier here than the shell method. The inner radius is 1. What is the
outer radius?

*Choice (b) is incorrect*

Hint: the disc method is easier here than the shell method. The inner radius is 1. What is the
outer radius?

*Choice (c) is incorrect*

Hint: the disc method is easier here than the shell method. The inner radius is 1. What is the
outer radius?

*Choice (d) is incorrect*

*Choice (e) is correct!*

The correct answer is
$\pi {\int}_{-1}^{1}\left({\left({x}^{3}+2\right)}^{2}-1\right)\phantom{\rule{1em}{0ex}}dx=\frac{44\pi}{7}.$ This
integral is set up using the disc method. Using the shell method, the integral is
$2\pi {\int}_{0}^{2}\left(y+1\right)\left(1-{\left(y-1\right)}^{\frac{1}{3}}\right)\phantom{\rule{1em}{0ex}}dy$, a
more difficult integral to evaluate.