Higher Maths
Integration

First we prepare for integration by writing each term in the form \(ax^n.\)

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

Then we integrate term-by-term, raising each power by \(1\) and dividing by the new power:

$$ \begin{eqnarray} I &=& \frac{4x^{\large\frac{3}{2}\normalsize}}{{}^{3}\!/\!_{2}}- \frac{3x^{-1}}{-1} +x +c \\[6pt] &=& \small\frac{2}{3}\normalsize\left( 4x^{\large\frac{3}{2}\normalsize} \right) +3x^{-1} +x +c \\[6pt] &=& \small\frac{8}{3}\normalsize x^{\large\frac{3}{2}\normalsize} + \small\frac{3}{x}\normalsize +x +c \\[6pt] &=& \small\frac{8}{3}\normalsize\sqrt{x^3} + \small\frac{3}{x}\normalsize +x +c \end{eqnarray} $$

Note that turning this answer into the same style as the question is not actually necessary. It has been shown here to encourage fuller understanding. However, a perfectly acceptable final answer would have been \(\frac{8}{3} x^{\large\frac{3}{2}\normalsize} +3x^{-1} +x +c\small.\)

The denominator consists of only one term, so we can prepare for differentiation by dividing each term in the numerator by the denominator:

$$ \begin{eqnarray} I &=& \int\small\frac{2x^4-5}{x^3}\normalsize\,dx \\[6pt] &=& \int\small\left(\frac{2x^4}{x^3}\normalsize-\small\frac{5}{x^3}\right)\normalsize\,dx \\[6pt] &=& \int\left(2x-5x^{-3}\right)\,dx \end{eqnarray} $$

Then we integrate term-by-term, raising each power by \(1\) and dividing by the new power:

$$ \begin{eqnarray} I &=& \small\frac{2x^2}{2}\normalsize-\small\frac{5x^{-2}}{-2}\normalsize+c \\[6pt] &=& x^2 + \small\frac{5}{2x^2}\normalsize +c \\ \end{eqnarray} $$

Once in integrable form, we integrate, substitute the upper and lower limits, and simplify:

$$ \begin{eqnarray} \int^{4}_{1}\frac{1}{3x^2}\,dx &=& \int^{4}_{1}\small\frac{1}{3}\,\normalsize x^{-2}\,dx\\[12pt] &=& \biggl[-\small\frac{1}{3}\normalsize x^{-1} \biggr]^4_1 \\[12pt] &=& -\!\small\frac{1}{3}\normalsize(4^{-1}) - \left(-\small\frac{1}{3}\normalsize (1^{-1})\right) \\[12pt] &=& -\!\small\frac{1}{3}\left(\small\frac{1}{4}\right) + \small\frac{1}{3}\normalsize \\[9pt] &=& -\!\small\frac{1}{12}\normalsize + \small\frac{1}{3}\normalsize \\[9pt] &=& -\!\small\frac{1}{12}\normalsize + \small\frac{4}{12}\normalsize \\[9pt] &=& \:\small\frac{3}{12}\normalsize \\[9pt] &=& \:\small\frac{1}{4}\normalsize \end{eqnarray} $$

Alternatively, we could have factorised the \(-\frac{1}{3}\) out of the square brackets, which would have simplified the fraction arithmetic:

$$ \begin{eqnarray} &\,& \!\!\!\!\! \biggl[-\small\frac{1}{3}\normalsize x^{-1}\biggr]^4_1 \\[9pt] &=& -\!\small\frac{1}{3}\normalsize\biggl[\small\frac{1}{x}\normalsize\biggr]^4_1 \\[9pt] &=& -\!\small\frac{1}{3}\normalsize\left(\small\frac{1}{4}\normalsize-1\right) \\[9pt] &=& -\!\small\frac{1}{3}\normalsize\left(-\small\frac{3}{4}\normalsize\right) \\[9pt] &=& \:\small\frac{1}{4}\normalsize \end{eqnarray} $$

This example is ready to integrate without any initial preparation.

$$ \begin{eqnarray} &\,& \!\!\!\!\int^{\sqrt{3}}_{\sqrt{2}}(4x^3-2x)\,dx \\[9pt] &=& \biggl[x^4-x^2\biggr]^{\sqrt{3}}_{\sqrt{2}} \\[9pt] &=& \left(\left(\sqrt{3}\right)^4-\left(\sqrt{3}\right)^2\right)-\left(\left(\sqrt{2}\right)^4-\left(\sqrt{2}\right)^2\right) \\[11pt] &=& \left(9-3\right)-\left(4-2\right) \\[9pt] &=& 6-2 \\[9pt] &=& 4 \end{eqnarray} $$

Here we use the chain rule for integration. This involves raising the power by \(1\) and dividing by the new power but also dividing by the derivative of the function in the brackets: in this case the derivative of \(2x+3,\) which is \(2.\)

$$ \begin{eqnarray} \int(2x+3)^5\,dx &=& \frac{(2x+3)^6}{(6)(2)}+c \\[8pt] &=& \small\frac{1}{12}\normalsize (2x+3)^6 + c \end{eqnarray} $$

This is very similar to the previous example, except that we must first rearrange it slightly.

$$ \begin{eqnarray} \int\frac{4}{(9-x)^6}\,dx &=& \int 4(9-x)^{-6}\,dx \\[9pt] &=& \small\frac{4(9-x)^{-5}}{(-5)(-1)}\normalsize+c \\[9pt] &=& \small\frac{4}{5}\normalsize (9-x)^{-5} +c \\[9pt] &=& \small\frac{4}{5(9-x)^5}\normalsize +c \end{eqnarray} $$

This example uses the chain rule for integration with the first of the given standard integrals, above. We need to divide by the derivative of the contents of the bracket, which is just the coefficient of \(x.\)

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

This tricky example of definite integration requires the chain rule for integration and a clear understanding of the exact values of trigonometric functions in the four quadrants.

Our advice is to work in radians throughout, if you possibly can.

$$ \begin{eqnarray} &\,& \!\!\!\!\!\!\int^{\large{\frac{\pi}{6}}}_{0}\,5\,\text{cos}\left(3x+\small\frac{\pi}{4}\normalsize\right)\,dx \\[10pt] &=& \biggl[\small\frac{5}{3}\normalsize\,\text{sin}\left(3x+\small\frac{\pi}{4}\normalsize\right)\biggr]^{\large\frac{\pi}{6}\normalsize}_{\large\,0\normalsize} \\[10pt] &=& \small\frac{5}{3}\normalsize\biggl[\,\text{sin}\left(3x+\small\frac{\pi}{4}\normalsize\right)\biggr]^{\large\frac{\pi}{6}\normalsize}_{\large\,0\normalsize} \\[10pt] &=& \small\frac{5}{3}\large\left(\normalsize \text{sin}\large\left(\normalsize \small\frac{\pi}{2}\normalsize+\small\frac{\pi}{4}\normalsize \large\right)\normalsize - \text{sin}\large\left(\normalsize 0+\small\frac{\pi}{4}\normalsize \large\right)\normalsize \large\right)\normalsize \\[10pt] &=& \small\frac{5}{3}\large\left(\normalsize \text{sin}\,\small\frac{3\pi}{4}\normalsize - \text{sin}\,\small\frac{\pi}{4}\normalsize\large\right)\normalsize \\[10pt] &=& \small\frac{5}{3}\left(\small\frac{1}{\sqrt{2}}\normalsize-\small\frac{1}{\sqrt{2}}\normalsize\right) \\[10pt] &=& \small\frac{5}{3}\normalsize\times 0 \\[10pt] &=& 0 \end{eqnarray} $$

This is an example of a differential equation.

We are given the derivative \(f'(x)\) and have to integrate to find the general solution.

We then use the additional condition to find the constant of integration \(c,\) which gives us the particular solution.

$$ \begin{eqnarray} f(x) &=& \int\frac{3x-2}{\sqrt{x}}\,dx \\[8pt] &=& \int\frac{3x-2}{x^{\large{}^{1}\!/\!_{2}\normalsize}}\,dx \\[8pt] &=& \int\left(\frac{3x}{x^{\large{}^{1}\!/\!_{2}\normalsize}}-\frac{2}{x^{\large{}^{1}\!/\!_{2}\normalsize}}\right)\,dx \\[8pt] &=& \int\left(3x^{\large{\frac{1}{2}}}-2x^{-\large{\frac{1}{2}}}\right)\,dx \\[8pt] &=& \small\frac{2}{3}\normalsize\left(3x^{\large{\frac{3}{2}}}\right)-2\left(2x^{\large{\frac{1}{2}}}\right)+c \\[8pt] &=& 2x^{\large{\frac{3}{2}}}-4x^{\large{\frac{1}{2}}}+c \end{eqnarray} $$

This is the general solution as it still involves the arbitrary constant of integration.

However, we also know that \(f(9)=45\) so we can substitute this to find the particular solution.

$$ \begin{gather} f(x)=2x^{\large{\frac{3}{2}}}-4x^{\large{\frac{1}{2}}}+c \\[9pt] f(9)=2(9^{\large{\frac{3}{2}}})-4(9^{\large{\frac{1}{2}}})+c \\[9pt] 45=2\left(\sqrt{9}\right)^3-4\sqrt{9}+c \\[9pt] 45=2(3^3)-4(3)+c \\[9pt] 45=54-12+c \\[9pt] 45=42+c \\[9pt] c=3 \end{gather} $$

So the particular solution is \(f(x) = 2x^{\large{\frac{3}{2}}}-4x^{\large{\frac{1}{2}}}+3\small.\)

This is another way of presenting a differential equation.

We integrate and then substitute the \(x\) and \(y\) values of the given point to find the particular solution.

$$ \begin{eqnarray} y &=& \int\left(6x^2+\small\frac{1}{x^2}\normalsize\right)\,dx \\[8pt] &=& \int\left(6x^2+x^{-2}\right)\,dx \\[8pt] &=& \:2x^3-x^{-1}+c \\[8pt] &=& \:2x^3-\small\frac{1}{x}\normalsize+c \end{eqnarray} $$

Now we substitute \(x=-1\) and \(y=3.\)

$$ \begin{gather} y=2x^3-\small\frac{1}{x}\normalsize +c \\[6pt] 3=2(-1)^3-\small\frac{1}{-1}\normalsize +c \\[6pt] 3=-2+1+c \\[6pt] 3=-1+c \\[6pt] c=4 \end{gather} $$

So the particular solution is \(y = 2x^3-\large\frac{1}{x}\normalsize+4\)

Note that the graph of \(y=3x^2-12\) is a parabola with a minimum turning point. So the enclosed area is below the \(x\)-axis.

We first need to find where the curve intersects the \(x\)-axis so that we know the limits of our definite integration. The equation of the \(x\)-axis is \(y=0\) so we substitute \(y=0\):

$$ \begin{gather} 3x^2-12=0\\[7pt] 3(x^2-4)=0\\[7pt] 3(x+2)(x-2)=0\\[7pt] x+2=0\:\:\small\textsf{or}\normalsize\:\:x-2=0\\[7pt] x=-2\:\:\small\textsf{or}\normalsize\:\:x=2\\[7pt] \end{gather} $$

Then we integrate between these limits:

$$ \begin{eqnarray} && \!\!\!\!\!\!\int^{2}_{-2}(3x^2-12)\,dx \\[9pt] &=& \biggl[x^3-12x\biggr]^{\,2}_{-2} \\[9pt] &=& \left(2^3-12(2)\right)-\left((-2)^3-12(-2)\right) \\[9pt] &=& \left(8-24\right)-\left(-8+24\right) \\[9pt] &=& -\!16-16 \\[9pt] &=& -\!32 \end{eqnarray} $$

The negative sign is because this area is below the \(x\)-axis. However, there is no such thing as a negative area, so the actual area is \(32\) units².

First we need to equate the functions to find the \(x\)-coordinates of the points of intersection:

$$ \begin{gather} -x^2+2x=3x-12 \\[7pt] -x^2-x+12=0 \\[7pt] (x+4)(-x+3)=0 \\[7pt] x+4=0\:\:\small\textsf{or}\normalsize\:\:-\!x+3=0\\[7pt] x=-4\:\:\small\textsf{or}\normalsize\:\:x=3\\[7pt] \end{gather} $$

Because the coefficient of \(x^2\) is negative, this is a parabola with a maximum turning point, so the straight line is below the required area and the parabola is above.

So now we integrate 'Upper – Lower' between \(-4\) and \(3\small.\)

$$ \begin{eqnarray} \textsf{Area} &=& \int^{3}_{-4}\left((-x^2+2x)-(3x-12)\right)\,dx \\[7pt] &=& \int^{3}_{-4}\left(-x^2-x+12\right)\,dx \\[7pt] &=& \biggl[-\small\frac{1}{3}\normalsize x^3-\small\frac{1}{2}\normalsize x^2+12x\biggr]^{\,3}_{-4} \\[7pt] &=& \left(-9-\small\frac{9}{2}\normalsize+36\right)-\left(\small\frac{64}{3}\normalsize-8-48\right) \\[7pt] &=& \left(27-\small\frac{9}{2}\normalsize\right)-\left(\small\frac{64}{3}\normalsize-56\right) \\[7pt] &=& \left(\small\frac{54}{2}\normalsize-\small\frac{9}{2}\normalsize\right)-\left(\small\frac{64}{3}\normalsize-\small\frac{168}{3}\normalsize\right) \\[7pt] &=& \small\frac{45}{2}\normalsize-\left(-\small\frac{104}{3}\normalsize\right) \\[7pt] &=& \small\frac{45}{2}\normalsize+\small\frac{104}{3}\normalsize \\[7pt] &=& \small\frac{135}{6}\normalsize+\small\frac{208}{6}\normalsize \\[7pt] &=& \small\frac{343}{6}\normalsize \\[4pt] \end{eqnarray} $$

So the area enclosed by the two graphs is \(\large\frac{343}{6}\normalsize\) units², or \(57\large\frac{1}{6}\normalsize\) units².

Note 1: The fraction work in this example is harder than anything that has ever appeared on the non-calculator paper. However, despite the fact that you are unlikely to face such awkward numbers, you do need to be able to manipulate fractions like this, avoiding sign errors. It can take quite a lot of practice!

Note 2: You may have noticed that you obtain the expression \(-x^2-x+12\) both when finding the roots and when simplifying 'upper minus lower'. That's no accident. With experience, you could use this to make the above working a little more efficient.

This 4-mark question requires some rearrangement before using the chain rule to integrate.

$$ \begin{eqnarray} &\,& \!\!\!\!\!\!\int\normalsize\frac{1}{(5-4x)^{\frac12}}\,dx \\[8pt] &=& \int\normalsize (5-4x)^{-\frac12}\,dx \\[8pt] &=& \frac{(5-4x)^{\frac12}}{({}^{1}\!/\!_{2})(-4)}+c \\[10pt] &=& -\!\small\frac{1}{2}\normalsize (5-4x)^{\frac12}+c \end{eqnarray} $$

This is a differential equation. We are given the derivative and have to integrate to find the general solution.

We then use the additional condition to find the constant of integration \(c,\) which gives us the particular solution.

$$ \begin{eqnarray} y &=& \int(6x^2-3x+4)\,dx \\[8pt] &=& 2x^3-\small\frac{3}{2}\normalsize x^2+4x+c \end{eqnarray} $$

This is the general solution as it still involves the arbitrary constant of integration.

However, we are also told that \(y=14\) when \(x=2\small,\) so we can substitute this to find the particular solution.

$$ \begin{eqnarray} y &=& 2x^3-\small\frac{3}{2}\normalsize x^2+4x+c \\[6pt] 14 &=& 2(2^3)-\small\frac{3}{2}\normalsize(2^2)+4(2)+c \\[6pt] 14 &=& 16-\small\frac{3}{2}\normalsize (4)+8+c \\[6pt] 14 &=& 16-6+8+c \\[6pt] 14 &=& 18+c \\[6pt] c &=& 14-18 \\[6pt] c &=& -\!4 \end{eqnarray} $$

So the particular solution is \(y=2x^3-\large\frac{3}{2}\normalsize x^2+4x-4\small.\)

This question requires the chain rule for integration and a clear understanding of the exact values of trig functions in the four quadrants. At one point in the working we need to evaluate \(sin\,(-\large\frac{\pi}{6})\small,\) so we need to understand that this is a quadrant 4 angle and that its sine is negative.

$$ \begin{eqnarray} &\,& \!\!\!\!\!\!\int^{\frac{\large\pi}{9}\normalsize}_{\small 0\normalsize}\normalsize\,\text{cos}\left(3x-\small\frac{\pi}{6}\normalsize\right) \,dx \\[10pt] &=& \biggl[\small\frac{1}{3}\normalsize\,\text{sin}\left(3x-\small\frac{\pi}{6}\normalsize\right)\biggr]^{\large\frac{\pi}{9}\normalsize}_{0} \\[10pt] &=& \small\frac{1}{3}\normalsize\,\text{sin}\left(\small\frac{3\pi}{9}\normalsize-\small\frac{\pi}{6}\normalsize \right)\,-\,\small\frac{1}{3}\normalsize\,\text{sin}\left(3(0)-\small\frac{\pi}{6}\normalsize\right) \\[10pt] &=& \small\frac{1}{3}\normalsize\,\text{sin}\left(\small\frac{\pi}{3}\normalsize-\small\frac{\pi}{6}\normalsize\right)\,-\,\small\frac{1}{3}\normalsize\,\text{sin}\left(-\small\frac{\pi}{6}\normalsize\right) \\[10pt] &=& \small\frac{1}{3}\normalsize\,\text{sin}\,\small\frac{\pi}{6}\normalsize\,-\,\small\frac{1}{3}\normalsize\,\text{sin}\left(-\small\frac{\pi}{6}\normalsize\right) \\[10pt] &=& \small\frac{1}{3}\normalsize\times\small\frac{1}{2}\normalsize\,-\,\small\frac{1}{3}\normalsize\times\left(-\small\frac{1}{2}\normalsize\right) \\[10pt] &=& \small\frac{1}{6}\normalsize + \small\frac{1}{6}\normalsize \\[10pt] &=& \small\frac{1}{3}\normalsize \\[10pt] \end{eqnarray} $$

First we prepare for integration by writing the first term in the form \(ax^n\small.\)

$$ \begin{eqnarray} I &=& \int\left(6\sqrt{x}-4x^{-3}+5\right)\,dx \\[6pt] &=& \int \left(6x^{\large1\over 2\normalsize}-4x^{-3}+5\right)\,dx \end{eqnarray} $$

Then we integrate, raising each power by \(1\) and dividing by the new power:

$$ \begin{eqnarray} I &=& \frac{6x^{\large\frac{3}{2}\normalsize}}{{}^{3}\!/\!_{2}}- \frac{4x^{-2}}{-2} +5x +c \\[6pt] &=& \small\frac{2}{3}\normalsize\left( 6x^{\large\frac{3}{2}\normalsize} \right) +2x^{-2} +5x +c \\[6pt] &=& 4x^{\large\frac{3}{2}\normalsize} + 2x^{-2} +5x +c \end{eqnarray} $$

This 4-mark definite integration requires the chain rule.

$$ \begin{eqnarray} &\,& \!\!\!\!\!\!\int^{2}_{-5}(10-3x)^{-\large\frac{1}{2}\normalsize}\,dx \\[9pt] &=& \Biggl[\frac{(10-3x)^{\large\frac{1}{2}\normalsize}}{({}^{1}\!/\!_{2})(-3)}\Biggr]^{\,2}_{-5} \\[11pt] &=& \Biggl[\frac{(10-3x)^{\large\frac{1}{2}\normalsize}}{-{}^{3}\!/\!_{2}}\Biggr]^{\,2}_{-5} \\[11pt] &=& -\!\small\frac{2}{3}\normalsize\biggl[(10-3x)^{\large\frac{1}{2}\normalsize}\biggl]^{\,2}_{-5} \\[10pt] &=& -\!\small\frac{2}{3}\normalsize\left((10-3\!\times\!2)^{\large\frac{1}{2}\normalsize}-(10-3\!\times\!-5)^{\large\frac{1}{2}\normalsize}\right) \\[10pt] &=& -\!\small\frac{2}{3}\normalsize\left(\sqrt{4}-\sqrt{25}\right) \\[9pt] &=& -\!\small\frac{2}{3}\normalsize\left(2-5\right) \\[9pt] &=& -\!\small\frac{2}{3}\normalsize\left(-3\right) \\[9pt] &=& \ \,2 \end{eqnarray} $$

First we prepare for integration by writing the second term in the form \(ax^n.\)

$$ \begin{eqnarray} I &=& \int\left(2x^5-6\sqrt{x}\,\right)\,dx \\[6pt] &=& \int\left(2x^5-6x^{\large1\over 2\normalsize}\right)\,dx \end{eqnarray} $$

Then we integrate each term, adding \(1\) to each power and dividing by the new power:

$$ \begin{eqnarray} I &=& \frac{2x^{6}}{6}-\frac{6x^{\large\frac{3}{2}\normalsize}}{{}^{3}\!/\!_{2}}+c \\[9pt] &=& \small\frac13\normalsize x^6-\small\frac{2}{3}\normalsize\left( 6x^{\large\frac{3}{2}\normalsize}\right)+c \\[9pt] &=& \small\frac13\normalsize x^6-4x^{\large\frac{3}{2}\normalsize}+c \end{eqnarray} $$

Note that there is no need to convert the second term back into surd form, but you may do so if you prefer.

This question requires the chain rule for integration.

Make sure your calculator is set to radians, not degrees!

$$ \begin{eqnarray} &\,& \!\!\!\!\!\int^{\large{\frac{\pi}{7}}}_{0}\text{sin}\,5x\,dx \\[9pt] &=& \biggl[-\small\frac{1}{5}\normalsize\,\text{cos}\,5x\,\biggr]^{\large\frac{\pi}{7}\normalsize}_{\,0\normalsize} \\[10pt] &=& -\small\frac{1}{5}\normalsize\,\text{cos}\,5\left(\small\frac{\pi}{7}\right)\,-\,\left(-\small\frac{1}{5}\normalsize\,\text{cos}\,5(0)\right) \\[10pt] &=& -\small\frac{1}{5}\normalsize\,\text{cos}\,\small\frac{5\pi}{7}\,\normalsize+\,\small\frac{1}{5}\normalsize\,\text{cos}\,0 \\[10pt] &=& -\small\frac{1}{5}\normalsize\left(-0.62348\cdots\right)+\small\frac{1}{5}\normalsize\left(1\right) \\[10pt] &\approx& \:0.3247 \end{eqnarray} $$

From the diagram, we can see that the parabola is above the cubic curve between \(x\!=\!0\) and \(x\!=\!2\small.\)

Let's simplify 'Upper – Lower' to start:

$$ \begin{eqnarray} && \!\!\!\!\!\!(6+4x-2x^2)-(x^3-6x^2+11x) \\[7pt] &=& 6+4x-2x^2-x^3+6x^2-11x \\[7pt] &=& -\!x^3+4x^2-7x+6 \\[7pt] \end{eqnarray} $$

Now we integrate 'Upper – Lower' between \(0\) and \(2\small.\)

$$ \begin{eqnarray} \textsf{Area} &=& \int^{2}_{0}\left(-x^3+4x^2-7x+6\right)\,dx \\[8pt] &=& \biggl[-\tfrac{1}{4}x^4+\tfrac{4}{3}x^3-\tfrac{7}{2}x^2+6x\biggr]^{\,2}_{\,0} \\[12pt] &=& \left(-\tfrac{1}{4}(2^4)+\tfrac{4}{3}(2^3)-\tfrac{7}{2}(2^2)+6(2)\right)-0 \\[12pt] &=& -\!\tfrac{1}{4}(16)+\tfrac{4}{3}(8)-\tfrac{7}{2}(4)+12 \\[12pt] &=& -\!4+\small\frac{32}{3}\normalsize-14+12 \\[10pt] &=& \small\frac{32}{3}\normalsize-6 \\[10pt] &=& \small\frac{32}{3}\normalsize-\small\frac{18}{3} \\[10pt] &=& \small\frac{14}{3}\normalsize \\[10pt] \end{eqnarray} $$

So the shaded area is \(\large\frac{14}{3}\normalsize\) units².

Note: You could, of course, use your calculator to help with the last few lines of working, and that would probably be a good idea to minimise the chance of error, but we just wanted to show that it isn't really necessary.

This is a differential equation. We are given the derivative and have to integrate to find the general solution.

We then use the additional condition to find the constant of integration \(c,\) which gives us the particular solution.

$$ \begin{eqnarray} y &=& \int(6\,\text{cos}\,x+8\,\text{sin}\,2x)\,dx \\[8pt] &=& 6\,\text{sin}\,x-4\,\text{cos}\,2x+c \end{eqnarray} $$

This is the general solution as it still involves the arbitrary constant of integration.

We are also told that \(y=4\) when \(x=\large\frac{\pi}{6}\small,\) so we substitute these values to obtain the particular solution.

$$ \begin{eqnarray} y &=& 6\,\text{sin}\,x-4\,\text{cos}\,2x+c \\[6pt] 4 &=& 6\,\text{sin}\,\small\frac{\pi}{6}\normalsize-4\,\text{cos}\,2\left(\small\frac{\pi}{6}\normalsize\right)+c \\[6pt] 4 &=& 6\,\text{sin}\,\small\frac{\pi}{6}\normalsize-4\,\text{cos}\,\small\frac{\pi}{3}\normalsize+c \\[6pt] 4 &=& 6\!\times\!\small\frac{1}{2}\normalsize-4\!\times\!\small\frac{1}{2}\normalsize+c \\[6pt] 4 &=& 3-2+c \\[6pt] 4 &=& 1+c \\[6pt] c &=& 3 \\[6pt] \end{eqnarray} $$

So the particular solution is \(y=6\,\text{sin}\,x-4\,\text{cos}\,2x+3\small.\)

The enclosed area is the answer to the definite integral with lower limit \(x\!=\!2\) and upper limit \(x\!=\!4\small.\)

$$ \begin{eqnarray} \textsf{Area} &=& \int^{4}_{2}\normalsize\left(x^2-2x+3\right)\,dx \\[9pt] &=& \biggl[ \frac{x^3}{3}-x^2+3x\biggr]^{\,4}_{\,2} \\[9pt] &=& \left(\small\frac{4^3}{3}\normalsize\!-\!4^2\!+\!3(4)\right)-\left(\small\frac{2^3}{3}\normalsize\!-\!2^2\!+\!3(2)\right)\\[9pt] &=& \left(\small\frac{64}{3}\normalsize-16+12\right)-\left(\small\frac{8}{3}\normalsize-4+6\right)\\[9pt] &=& \left(\small\frac{64}{3}\normalsize-4\right)-\left(\small\frac{8}{3}\normalsize+2\right)\\[9pt] &=& \:\small\frac{64}{3}\normalsize-4-\small\frac{8}{3}\normalsize-2\\[9pt] &=& \:\small\frac{56}{3}\normalsize-6\\[9pt] &=& \:\small\frac{56}{3}\normalsize-\small\frac{18}{3}\\[9pt] &=& \:\small\frac{38}{3}\\[9pt] \end{eqnarray} $$

So the shaded area is \(\large\frac{38}{3}\normalsize\) units².

Note: You could, of course, use your calculator to help with the last few lines of working, and that would probably be a good idea to reduce the likelihood of error, but it isn't really necessary.

This simple 2-mark question requires the the chain rule for integration.

So, as usual, we raise the power by \(1\) and divide by the new power, but we also divide by the derivative of the 'inner function'.

$$ \begin{eqnarray} \int(3x+2)^7\,dx &=& \frac{(3x+2)^8}{(8)(3)}+c \\[8pt] &=& \frac{(3x+2)^8}{24}+c \\[8pt] \end{eqnarray} $$