Advanced Higher Maths
Integration

The key here is to notice that the numerator is very closely related to the derivative of the denominator:

$$ \small\frac{d}{dx}\normalsize\left(2x^3-2x+1\right)=6x^2-2 $$

So the numerator is exactly half of the derivative of the denominator. Therefore:

$$ \displaystyle\int\!\small\frac{3x^2-1}{2x^3-2x+1}\normalsize\,dx= \small\frac{1}{2}\normalsize\,\text{ln}\,\vert 2x^3\!-\!2x+1\vert +c $$

Note: Check for yourself that this answer differentiates back to the question, using the chain rule.

Video solution by Clelland Maths 

This solution requires the following standard integral from the formulae list :

$$ \displaystyle\int\normalsize\!\!\small\frac{dx}{\sqrt{a^2-x^2}}\normalsize=\text{sin}^{-1}\left(\small\frac{x}{a}\normalsize\right)+c $$

Here, \(a\!=\!2\small,\) we have \(\raise 0.2pt{(3x)^2}\) instead of \(\raise 0.2pt{x^2}\) and the entire standard integral is multiplied by \(6\small.\)

So, making use of the chain rule for integration:

$$ \begin{eqnarray} \displaystyle\int\normalsize\!\small\frac{6\,dx}{\sqrt{4-9x^2}} &=& \small\frac{6}{3}\normalsize\,\text{sin}^{-1}\small\left(\frac{3x}{2}\right)\normalsize+c \\[6pt] &=& 2\,\text{sin}^{-1}\left(\small\frac{3x}{2}\normalsize\right)+c \\[6pt] \end{eqnarray} $$

Video solution by Clelland Maths 

\(x^2-2x-15\) factorises as \((x-5)(x+3)\) so:

$$ \frac{3x-7}{(x-5)(x+3)} = \frac{A}{x-5} + \frac{B}{x+3} $$

$$ 3x-7 = A(x+3) + B(x-5) $$

Substituting well-chosen values of \(\raise 0.2pt{x}\) lets us find the values of \(\raise 0.2pt{A}\) and \(\raise 0.2pt{B\small.}\)

\(x=-3\implies -16=B(-8)\implies B=2\)
\(x=5\implies 8=A(8)\implies A=1\)

So now we are ready to integrate:

$$ \begin{eqnarray} && \displaystyle\int\normalsize\frac{3x-7}{x^2-2x-15}\normalsize\,dx \\[6pt] &=& \displaystyle\int\normalsize\left(\frac{1}{x-5}+\frac{2}{x+3}\right)\,dx \\[12pt] &=& \text{ln}\,\vert x-5\vert + 2\,\text{ln}\,\vert x+3\vert+c \\[6pt] \end{eqnarray} $$

Note: If you wanted to present your final answer in the form \(\text{ln}\,\vert (x-5)(x+3)^2\vert+c\small,\) that would also be correct, but we prefer the more 'linear' style of the answer, as shown above.

Video solution by Clelland Maths 

This example also requires partial fractions, but this time the denominator contains a repeated factor, so the form of the partial fractions is:

$$ \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{2x-1} $$

Multiplying through by the denominator gives:

$$ x\!\small+\normalsize\!4=A(x\!\small+\normalsize\!1)(2x\!\small-\normalsize\!1)+B(2x\!\small-\normalsize\!1)+C(x\!\small+\normalsize\!1)^2 $$

Now we can substitute well-chosen values of \(\raise 0.2pt{x}\) to find two of our three unknowns:

\(x=-1\implies 3=B(-3)\implies B=-1\)
\(x=\large\frac{1}{2}\normalsize\implies\large\frac{9}{2}\normalsize=C\left(\large\frac{9}{4}\normalsize\right)\implies C=2\)

No substitution will let us find \(A\small.\) Instead, we can equate the constant terms:

$$ \begin{gather} 4=A(-1)+B(-1)+C(1)\\[6pt] 4=A(-1)+(-1)(-1)+2(1)\\[6pt] 4=-A+1+2\\[6pt] A=-1\\[6pt] \end{gather} $$

So now we can integrate and evaluate:

$$ \begin{eqnarray} && \displaystyle\int^{2}_{1}\left(-\frac{1}{x+1}-\frac{1}{(x+1)^2}+\frac{2}{2x-1}\right)\,dx\\[9pt] && \displaystyle\int^{2}_{1}\left(-\frac{1}{x+1}-(x+1)^{-2}+\frac{2}{2x-1}\right)\,dx\\[9pt] &=& \bigg[\normalsize\!-\text{ln}\,\vert x+1\vert+(x+1)^{-1}+\text{ln}\,\vert 2x-1\vert\,\bigg]^{2}_{1}\\[9pt] &=& \left(-\text{ln}\,3+\small\frac{1}{3}\normalsize+\text{ln}\,3\right)-\left(-\text{ln}\,2+\small\frac{1}{2}\normalsize+\cancelto{0}{\text{ln}\,1}\right)\\[9pt] &=& \cancel{-\text{ln}\,3}+\small\frac{1}{3}\normalsize+\cancel{\text{ln}\,3}+\text{ln}\,2-\small\frac{1}{2}\normalsize\\[9pt] &=& \:\text{ln}\,2-\small\frac{1}{6}\\[6pt] \end{eqnarray} $$

Video solution by Clelland Maths 

We start by differentiating \(u\) to obtain an expression for \(dx\small.\)

$$ \begin{eqnarray} u &=& \text{tan}\,x \\[6pt] \small\frac{du}{dx}\normalsize &=& \text{sec}^{2}\,x \\[6pt] du &=& \text{sec}^{2}\,x\,dx \\[6pt] dx &=& \small\frac{du}{\text{sec}^{2}\,x} \end{eqnarray} $$

Now we subtitute this into the original integration to express it in terms of \(\raise 0.2pt{u\small.}\)

$$ \begin{eqnarray} \displaystyle\int\normalsize\small\frac{dx}{\text{sin}\,x\,\text{cos}\,x} &=& \small\displaystyle\int\frac{1}{\text{sin}\,x\,\text{cos}\,x}\normalsize\,.\,\small\frac{du}{\text{sec}^{2}\,x} \\[12pt] &=& \displaystyle\int\small\frac{\text{cos}^{2}\,x}{\text{sin}\,x\,\text{cos}\,x}\,du \\[12pt] &=& \displaystyle\small\int\frac{\text{cos}\,x}{\text{sin}\,x}\,du \\[12pt] &=& \displaystyle\small\int\frac{1}{\text{tan}\,x}\,du \\[12pt] &=& \displaystyle\small\int\frac{1}{u}\,du \\[10pt] &=& \ \text{ln}\,\vert u\vert+c \\[10pt] &=& \ \text{ln}\,\vert \text{tan}\,x\vert+c \\[6pt] \end{eqnarray} $$

Video solution by Clelland Maths 

Integration by parts could be used for this question, rather than substitution. Try it if you wish!

In general, though, you would be well advised to ignore the words 'or otherwise'.

We start by differentiating \(\raise 0.2pt{u}\small.\)

$$ \begin{gather} u=sin\,\theta \\[6pt] \small\frac{du}{d\theta}\normalsize=cos\,\theta \\[6pt] \end{gather} $$

Because this is a definite integration, we need to express the upper and lower limits in terms of the new variable.

\(\theta=\large\frac{\pi}{2}\normalsize\implies u=sin\,\large\frac{\pi}{2}\normalsize=1\)
\(\theta=\large\frac{\pi}{6}\normalsize\implies u=sin\,\large\frac{\pi}{6}\normalsize=\large\frac{1}{2}\)

Now we express the original integration in terms of \(\raise 0.2pt{u\small.}\)

$$ \begin{eqnarray} \int^{\large{\frac{\pi}{2}}}_{\large{\frac{\pi}{6}}}2\,sin^4\,\theta\,cos\,\theta\,d\theta &=& \int^{1}_{\large\frac12}2\,u^4\,\small\frac{du}{d\theta}\normalsize\,d\theta \\[12pt] &=& 2\,\int^{1}_{\large\frac12}u^4\,du \\[12pt] &=& 2\,\biggl[\small\frac{1}{5}\normalsize u^5 \biggr]^1_{\large\frac12} \\[12pt] &=& \small\frac{2}{5}\normalsize \,\biggl[ u^5 \biggr]^{1}_{\large\frac12} \\[12pt] &=& \small\frac{2}{5}\normalsize\,\left(1^5-\small\left(\frac{1}{2}\right)^{\!5}\normalsize\right) \\[12pt] &=& \small\frac{2}{5}\normalsize\,\left(1-\small\frac{1}{32}\normalsize\right) \\[10pt] &=& \small\frac{2}{5}\normalsize\,\left(\small\frac{31}{32}\normalsize\right) \\[10pt] &=& \small\frac{31}{80}\normalsize \end{eqnarray} $$

Note: You may have noticed that we took a slightly different approach in the top part of this solution than we did in example 5 above. In example 5, we obtained an expression for \(dx\) before switching variables, whereas here, we could see that \(cos\,\theta\) is a factor in the integrand, so we just wrote it as \(\large\frac{du}{d\theta}\) and then used the fact that \(\large\frac{du}{d\theta}\normalsize.\,d\theta=du\) to eliminate \(\theta\small.\) If that seems a bit convoluted to you, just stick to the method shown in example 5.

Video solution by Clelland Maths 

This question is a simple 3-mark single use of integration by parts:

$$ \int uv'\,dx=uv-\int u'v\,dx $$

First we examine the factors to decide which we should differentiate \((u)\) and which we should integrate \((v')\small.\)

As \(\text{ln}\,x\) differentiates to \(\large\frac{1}{x}\small,\) choosing \(\text{ln}\,x\) as \(u\) is the only way to go.

$$ \begin{matrix} u=\text{ln}\,x \:&\: v'=x^4 \\[5pt] u'=\large\frac{1}{x} \:&\: v=\large\frac{1}{5}\normalsize x^5 \\ \end{matrix} $$

Now we use integration by parts:

$$ \begin{eqnarray} \int \left(\text{ln}\,x\right)x^4\,dx &=& uv-\int u'v\,dx \\[9pt] &=& \left(\text{ln}\,x\right)\small\frac{1}{5}\normalsize x^5-\int \small\frac{1}{x}\normalsize\,.\,\small\frac{1}{5}\normalsize x^5\,dx \\[9pt] &=& \small\frac{1}{5}\normalsize x^5\,\text{ln}\,x-\small\frac{1}{5}\normalsize\int x^4\,dx \\[9pt] &=& \small\frac{1}{5}\normalsize x^5\,\text{ln}\,x-\small\frac{1}{5}\normalsize\,.\,\small\frac{1}{5}\normalsize x^5+c \\[9pt] &=& \small\frac{1}{5}\normalsize x^5\,\text{ln}\,x-\small\frac{1}{25}\normalsize x^5+c \end{eqnarray} $$

Video solution by Clelland Maths 

This 5-mark question will require us to use integration by parts twice, as you are about to see.

First we examine the factors to decide which we should differentiate \((u)\) and which we should integrate \((v')\small.\)

The exponential factor would work either way, so we consider the \(x^2\) factor. Differentiating it will reduce its exponent, so we will differentiate \(x^2\) and integrate \(e^{3x}\small.\)

$$ \begin{matrix} u=x^2 \:&\: v'=e^{3x} \\[5pt] u'=2x \:&\: v=\large\frac{1}{3}\normalsize e^{3x}\\ \end{matrix} $$

For brevity, let \(\text{I}=\displaystyle\int\normalsize\!x^2\,e^{3x}\,dx\small.\)

Now we apply integration by parts:

$$ \begin{eqnarray} \text{I} &=& uv-\int u'v\,dx \\[6pt] &=& x^2\,.\,\small\frac{1}{3}\normalsize e^{3x}-\int 2x\,.\,\small\frac{1}{3}\normalsize e^{3x}\,dx \\[6pt] &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{3}\normalsize\int x\,e^{3x}\,dx \end{eqnarray} $$

Now we need to integrate by parts again. It is important to choose our new \(u\) and new \(v'\) consistently with the first part, so that the exponent reduces again. So:

$$ \begin{matrix} u=x \:&\: v'=e^{3x} \\[5pt] u'=1 \:&\: v=\large\frac{1}{3}\normalsize e^{3x} \\ \end{matrix} $$

$$ \begin{eqnarray} \text{I} &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{3}\normalsize\left(x\,.\,\small\frac{1}{3}\normalsize e^{3x}-\!\!\int\!1\,.\small\frac{1}{3}\normalsize e^{3x}\,dx\right) \\[6pt] &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{9}\normalsize\,x\,e^{3x}+\small\frac{2}{9}\normalsize\int\!e^{3x}\,dx \\[6pt] &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{9}\normalsize\,x\,e^{3x}+\small\frac{2}{27}\normalsize e^{3x}+c \end{eqnarray} $$

Video solution by Clelland Maths 

First we examine the factors to decide which we should differentiate \((u)\) and which we should integrate \((v')\small.\) In this example, it doesn't matter.

We will differentiate the trigonometric factor and integrate the exponential factor in the solution below. You might want to try it the other way around, as an exercise. The answer should, of course, be the same either way.

$$ \begin{matrix} u=\text{cos}\,x \:&\: v'=e^x \\[5pt] u'=-\text{sin}\,x \:&\: v=e^x\\ \end{matrix} $$

For brevity, let \(\text{I}=\displaystyle\int\normalsize\!e^x\,\text{cos}\,x\,dx\small.\)

So now we apply integration by parts:

$$ \begin{eqnarray} \text{I} &=& uv-\int u'v\,dx \\[6pt] &=& (\text{cos}\,x)\,e^x-\!\int (-\text{sin}\,x)\,e^x\,dx \\[6pt] &=& e^x\,\text{cos}\,x+\!\int e^x\,\text{sin}\,x\,dx \\[6pt] \end{eqnarray} $$

So we have to integrate by parts again, this time using the following repurposed definitions of \(u\) and \(v'\):

$$ \begin{matrix} u=\text{sin}\,x \:&\: v'=e^x \\[5pt] u'=\text{cos}\,x \:&\: v=e^x \\ \end{matrix} $$

What is going to happen here is that the original integral will appear again on the right, enabling solution:

$$ \begin{eqnarray} \text{I} &=& e^x\,cos\,x+\left((\text{sin}\,x)\,e^x-\!\small\int\normalsize(\text{cos}\,x)\,e^x\,dx\right) \\[9pt] &=& e^x\,\text{cos}\,x+e^x\,\text{sin}\,x-\!\small\int\normalsize e^x\,\text{cos}\,x\,dx \\[9pt] &=& e^x\,\text{cos}\,x+e^x\,\text{sin}\,x-\text{I} \\[9pt] \end{eqnarray} $$

So all that remains is to add \(\text{I}\) to both sides, halve and not forget the arbitary constant of integration:

$$ \begin{gather} 2\text{I}=e^x\,\text{cos}\,x+e^x\,\text{sin}\,x+c\\[9pt] \text{I}=\small\frac{1}{2}\,\normalsize e^x\bigl(\text{cos}\,x+\text{sin}\,x\bigr)+c\\[9pt] \end{gather} $$

Video solution by Clelland Maths 

We are being asked to integrate a product, so this example requires integration by parts:

$$ \int\,uv'\,dx = uv-\!\int u'v\,dx $$

We start by examining the factors to decide which we should differentiate \((u)\) and which we should integrate \((v')\small.\)

Differentiating \(x^7\) would only reduce its exponent by one each time, so that's far from ideal. Moreover, you probably haven't yet seen a standard integral for \(\text{ln}\,x\small,\) let alone \((\text{ln}\,x)^{2}\small.\)

So we will differentiate \((\text{ln}\,x)^2\small\) and hope that it simplifies reasonably quickly.

$$ \begin{matrix} u=(\text{ln}\,x)^2 \:&\: v'=x^7 \\[5pt] u'=2(\text{ln}\,x)\left(\large\frac{1}{x}\right) \:&\: v=\large\frac{1}{8}\normalsize x^8 \\ \end{matrix} $$

For brevity, let \(\text{I}=\displaystyle\int\normalsize\!x^{7}\,(\text{ln}\,x)^{2}\,dx\small.\)

So now we apply integration by parts:

$$ \begin{eqnarray} \text{I} &=& (\text{ln}\,x)^2\,.\,\small\frac{1}{8}\normalsize x^8-2\int(\text{ln}\,x)\left(\small\frac{1}{x}\normalsize\right)\,.\,\small\frac{1}{8}\normalsize x^{8}\,dx \\[6pt] &=& \small\frac{1}{8}\normalsize x^8(\text{ln}\,x)^2-\small\frac{1}{4}\normalsize \int x^{7}\,\text{ln}\,x\,dx \\[6pt] &=& \small\frac{1}{8}\normalsize x^8(\text{ln}\,x)^2-\small\frac{1}{4}\,\normalsize\text{J}\,\textsf{, say.} \\[6pt] \end{eqnarray} $$

Integrating what we have called \(\text{J}\) needs a second application of integration by parts.

So, recycling the \(u,\,v\) notation:

$$ \begin{matrix} u=\text{ln}\,x \:&\: v'=x^7\\[5pt] u'=\large\frac{1}{x}\normalsize \:&\: v=\large\frac{1}{8}\normalsize x^8 \\ \end{matrix} $$

You will notice that \(u'\) is much simpler this time. That's a positive sign!

$$ \begin{eqnarray} \text{J} &=& (\text{ln}\,x)\,.\,\small\frac{1}{8}\normalsize x^8\,-\int\small\frac{1}{x}\normalsize\,.\,\small\frac{1}{8}\normalsize x^{8}\,dx \\[8pt] &=& \small\frac{1}{8}\normalsize x^{8}\,\text{ln}\,x-\small\frac{1}{8}\normalsize \int x^7\,dx \\[8pt] &=& \small\frac{1}{8}\normalsize x^{8}\,\text{ln}\,x-\small\frac{1}{8}\normalsize \left(\small\frac{1}{8}\normalsize x^8\right)+c \\[8pt] &=& \small\frac{1}{8}\normalsize x^{8}\,\text{ln}\,x-\small\frac{1}{64}\normalsize x^8+c \\[8pt] \end{eqnarray} $$

Subsituting \(\text{J}\) back into \(\text{I}\small,\) we obtain:

$$ \begin{eqnarray} \text{I} &=& \small\frac{1}{8}\normalsize x^8(\text{ln}\,x)^2-\small\frac{1}{4}\normalsize \left(\small\frac{1}{8}\normalsize x^{8}\,\text{ln}\,x-\small\frac{1}{64}\normalsize x^{8}\right)+c \\[8pt] &=& \small\frac{1}{8}\normalsize x^8(\text{ln}\,x)^2-\small\frac{1}{32}\normalsize x^{8}\,\text{ln}\,x+\small\frac{1}{256}\normalsize x^{8}+c \\[6pt] \end{eqnarray} $$

Video solution by Clelland Maths 

From Higher, the equation of a circle radius \(r\) centred on the origin is \(x^2+y^2=r^2\small.\)

Rearranging, we obtain \(y^2=r^2-x^2\small.\)

\(y=\sqrt{r^2-x^2}\) is the upper semicircle and \(y=-\sqrt{r^2-x^2}\) is the lower semicircle. We could rotate either of these through \(2\pi\) radians to form a sphere. In practice, we don't need to choose, as the volume requires \(y^2\small,\) not \(y\small.\)

Note that the lower limit of \(x\) is \(-r\) and the upper limit of \(x\) is \(r\small.\)

$$ \begin{eqnarray} V &=& \int^{r}_{-r}\pi y^{2}\,dx \\[9pt] &=& \pi\!\int^{r}_{-r}(r^2-x^2)\,dx \\[9pt] &=& \pi\,\biggl[r^{2}x-\small\frac{1}{3}\normalsize x^3\biggr]^{\phantom{.}r}_{-r} \\[9pt] &=& \pi\,\biggl[\left(r^{3}-\small\frac{1}{3}\normalsize r^3\right)-\left(-r^3+\small\frac{1}{3}\normalsize r^3\right)\biggr] \\[9pt] &=& \pi\left(r^{3}-\small\frac{1}{3}\normalsize r^3+r^3-\small\frac{1}{3}\normalsize r^3\right) \\[9pt] &=& \pi\left(2r^{3}-\small\frac{2}{3}\normalsize r^3\right) \\[9pt] &=& \pi\left(\small\frac{6}{3}\normalsize r^{3}-\small\frac{2}{3}\normalsize r^3\right) \\[9pt] &=& \small\frac{4}{3}\normalsize \pi r^3\ \small\textsf{cubic units} \\[9pt] \end{eqnarray} $$

Note: It is unlikely that a question like this would appear on an exam paper without prompting you to rotate a semicircle centred on the origin. See example 12 below for an actual exam question of this type.

Video solution by Clelland Maths 

In this question, the curve is to be rotated around the \(y\)-axis, not the \(x\)-axis.

So the required volume is given by:

where \(L\) and \(U\) are the lower and upper limits. So we need to find these limits and an expression for \(x^{2}\small.\)

As the limits are \(y\)-values, we solve for \(x\!=\!0\small.\)

$$ \begin{gather} 4x^2+9y^2=36\\[6pt] 4(0)^2+9y^2=36 \\[6pt] 9y^2=36 \\[6pt] y^2=4 \\[6pt] y=\pm\,2 \end{gather} $$

However, we are only rotating the upper-right quadrant of this curve, so the lower limit is \(0\small,\) not \(-2\small.\)

Now we rearrange to obtain an expression for \(x^{2}\small.\)

$$ \begin{gather} 4x^2+9y^2=36\\[6pt] 4x^2=36-9y^2 \\[6pt] x^2=\small\frac{1}{4}\normalsize(36-9y^2) \end{gather} $$

We are now ready to integrate to find the volume of revolution.

$$ \begin{eqnarray} V &=& \int^{2}_{0}\pi x^{2}\,dy \\[8pt] &=& \small\frac{\pi}{4}\normalsize\int^{2}_{0}(36-9y^2)\,dy \\[8pt] &=& \small\frac{\pi}{4}\normalsize\biggl[36y-\small\frac{9}{3}\normalsize y^3\biggr]^{2}_{0} \\[10pt] &=& \small\frac{\pi}{4}\normalsize\bigl[36y-3y^3\bigr]^{2}_{0} \\[10pt] &=& \small\frac{\pi}{4}\normalsize\bigl\{\left(36(2)\!-\!3(2^3)\right)-\left(36(0)\!-\!3(0^3)\right)\bigr\} \\[10pt] &=& \small\frac{\pi}{4}\normalsize\bigl(72-3(8)\bigr) \\[10pt] &=& \small\frac{\pi}{4}\normalsize\left(48\right) \\[10pt] &=& 12\pi\ \small\textsf{cubic units} \end{eqnarray} $$

Video solution by Clelland Maths 

(a)  This 5-mark part is quite a fiddly double integration by parts.

We start by examining the two factors and asking ourselves which we should differentiate \((u)\) and which we should integrate \((v')\small.\) It doesn't matter for \(e^{4x}\) but the quadratic should be differentiated to reduce its degree.

So now we apply integration by parts, twice:

$$ \begin{eqnarray} && \!\!\!\!\!\! \int^{\small{1}}_{\small{0}}\underbrace{(x^2-2x+1)}_{\textsf{u}}\underbrace{e^{4x}}_{\textsf{v'}}\,dx\\[8pt] &=& \Biggl[\small\frac{1}{4}\normalsize e^{4x}\left(x^2\!-\!2x\!+\!1\right)\Biggr]^{1}_{0}-\int^{1}_{0}\!\small\frac14\normalsize e^{4x}(2x\!-\!2)\,dx\\[8pt] &=& \left(0-\small\frac14\right)\normalsize-\small\frac12\normalsize\int^{\small{1}}_{\small{0}}\underbrace{(x\!-\!1)}_{\textsf{u}}\underbrace{e^{4x}}_{\textsf{v'}}\,dx\\[8pt] &=& -\!\small\frac14-\small\frac12\normalsize\Biggl[\small\frac{1}{4}\normalsize e^{4x}(x\!-\!1)- \int^{\small{1}}_{\small{0}}1.\small\frac{1}{4}\normalsize e^{4x}\,dx\Biggr]^{1}_{0}\\[10pt] &=& -\!\small\frac14-\small\frac12\normalsize\Biggl[\small\frac{1}{4}\normalsize e^{4x}(x\!-\!1)-\small\frac{1}{16}\normalsize e^{4x}\Biggr]^{1}_{0} \\[10pt] &=& -\!\small\frac14-\small\frac12\normalsize\left(\small\left(\normalsize0-\small\frac{1}{16}\normalsize e^4\small\right)\normalsize-\small\left(\normalsize-\small\frac14\normalsize-\small\frac{1}{16}\normalsize\small\right)\normalsize\right)\\[10pt] &=& -\!\small\frac14-\small\frac12\normalsize\left(-\small\frac{1}{16}\normalsize e^4+\small\frac{5}{16}\normalsize\right)\\[10pt] &=& -\!\small\frac14+\small\frac{1}{32}\normalsize e^4-\small\frac{5}{32}\normalsize\\[10pt] &=& \ \small\frac{1}{32}\normalsize e^4-\small\frac{13}{32}\normalsize\\[10pt] &=& \ \small\frac{1}{32}\normalsize\left(e^4-13\right)\\[10pt] \end{eqnarray} $$

(b)  This 3-mark part makes use of the answer to part (a) to shortcut the integration process.

$$ \begin{eqnarray} V &=& \int^{1}_{0}\pi y^{2}\,dx \\[8pt] &=& \pi\!\int^{1}_{0}\bigl\{\,4(x\!-\!1)\,e^{2x}\,\bigr\}^2\,dx \\[8pt] &=& 16\pi\!\int^{1}_{0}\bigl\{\,(x^2\!-\!2x\!+\!1)\,e^{4x}\,\bigr\}^2\,dx \\[10pt] &=& \small\frac{16\pi}{32}\normalsize\left(e^4-13\right)\\[10pt] &=& \small\frac{\pi}{2}\normalsize\left(e^4-13\right)\ \small\textsf{cubic units} \end{eqnarray} $$

Video solution by Clelland Maths