Calculating Areas (Edexcel IGCSE Further Pure Maths): Revision Note

• Be careful when substituting in to find $\mathrm{f}\left(b\right)-\mathrm{f}\left(a\right)$

  • It's quite easy to make mistakes here

  • Especially when fractions and negative numbers are involved

• Your calculator may be able to find the value of definite integrals

  • You can use this to check your work

• Look out for phrases in exam questions like "Use algebraic integration" or "Using calculus"

  • These mean that full working out of the integral 'by hand' is required

  • A calculator answer without working would not score marks

Show that

${\int }_2^{4}3x(x^2-2)\mathrm{d}x=144$

Start by expanding the brackets inside the integral

${\int }_2^4\left(3x^3-6x\right)\mathrm{d}x$

Integrate as usual (here it's a 'powers of $x$' integration)

Write the answer in square brackets with the integration limits outside

$\begin{array}{rcl}{\int }_2^4\left(3x^3-6x\right)\mathrm{d}x& =& {\left[3\left(\frac{x^{3+1}}{3+1}\right)-6\left(\frac{x^{1+1}}{1+1}\right)\right]}_2^4\\ & =& {\left[\frac{3}{4}{x}^4-3x^2\right]}_2^4\end{array}$

Now substitute 4 into that function
And subtract from it the function with 2 substituted in

$\begin{array}{rcl}{\left[\frac{3}{4}{x}^4-3x^2\right]}_2^4& =& \left(\frac{3}{4}{\left(4\right)}^4-3{\left(4\right)}^2\right)-\left(\frac{3}{4}{\left(2\right)}^4-3\left(2^2\right)\right)\\ & =& \left(192-48\right)-\left(12-12\right)\\ & =& 144-0\\ & =& 144\end{array}$

And that's the answer we were asked to show!

${\mathbf{\int }}_{\mathbf{2}}^{\mathbf{4}}\mathbf{3}\mathit{x}\left(x^2-2\right)\mathbf{d}\mathit{x}\mathbf{=}\mathbf{144}$

• Look out for questions that ask you to find an indefinite integral in one part

  • where '$+c$'  is needed in the answer

  • then in a later part use the same integral as a definite integral

    • where '$+c$' is not needed

• Add information to any diagram provided in the question

  • axes intercepts

  • values of limits

  • mark and shade the area you’re trying to find

• If no diagram is provided, sketch one!

The following diagram shows a part of the graph of the curve with equation  $y=3+2x-x^2$.  The region labelled $R$ is bounded by the curve, the positive $y$-axis, and the positive $x$-axis.

Find the exact area of region $R$.

Start by finding the integration limits.

The $y$-axis is the left boundary, so $x=0$ will be the lower integration limit

The right boundary is where the curve intercepts the $x$-axis
Solve the equation  $y=0$  to find its $x$-coordinate

$\begin{array}{rcl}3+2x-x^2& =& 0\\ x^2-2x-3& =& 0\\ \left(x+1\right)\left(x-3\right)& =& 0\end{array}$

$x=-1\mathrm{or}x=3$

We want the positive intercept, so the upper integral limit will be $x=3$

Put all that info into the definite integral

$\begin{array}{rcl}\mathrm{Area}& =& {\int }_0^3\left(3+2x-x^2\right)\mathrm{d}x\\ & =& {\left[3x+2\left(\frac{x^{1+1}}{1+1}\right)-\left(\frac{x^{2+1}}{2+1}\right)\right]}_0^3\\ & =& {\left[3x+x^2-\frac{1}{3}{x}^3\right]}_0^3\\ & =& \left(3\left(3\right)+{\left(3\right)}^2-\frac{1}{3}{\left(3\right)}^3\right)-\left(3\left(0\right)+{\left(0\right)}^2-\frac{1}{3}{\left(0\right)}^3\right)\\ & =& \left(9+9-9\right)-\left(0+0-0\right)\\ & =& 9-0\\ & =& 9\end{array}$

$\mathbf{9}{\mathbf{units}}^{\mathbf{2}}$

• If no diagram is provided, sketch one

  • This lets you see where the curve is above and below the $x$-axis

  • Then you can split up your integrals accordingly

The diagram below shows the graph of $y=\mathrm{f}\left(x\right)$, where  $\mathrm{f}\left(x\right)=(x+4)(x-1)(x-5)$.

The region$R_1$ is bounded by the curve $y=\mathrm{f}\left(x\right)$ and the positive $x$- and $y$-axes.

The region$R_2$ is bounded by the curve $y=\mathrm{f}\left(x\right)$, the positive $x$-axis, and the line $x=3$.

(a) Determine the coordinates of the point labelled$P$.

That point is one of the $x$-axis intercepts  of the function

Solve $\mathrm{f}\left(x\right)=0$ to find what those area

$\left(x+4\right)\left(x-1\right)\left(x-5\right)=0$

$x=-4,1,5$

It has to be positive, because the point is to the right of the origin
And it has to be less than 3, because it's to the left of the line $x=3$
So it must be the intercept at $x=1$

$\left(1,0\right)$

(b) Find the exact total area of the shaded regions $R_1$ and $R_2$.

$R_1$ is the region between $x=0$ and $x=1$

It lies totally above the $x$-axis, so the definite integral will calculate the area directly

$\begin{array}{rcl}\mathrm{Area}\mathrm{of}{R}_1& =& {\int }_0^1\left(x+4\right)\left(x-1\right)\left(x-5\right)\mathrm{d}x\\ & =& {\int }_0^1\left(x^3-2x^2-19x+20\right)\mathrm{d}x\\ & =& {\left[\frac{1}{4}{x}^4-\frac{2}{3}{x}^3-\frac{19}{2}{x}^2+20x\right]}_0^1\\ & =& \left(\frac{1}{4}{\left(1\right)}^4-\frac{2}{3}{\left(1\right)}^3-\frac{19}{2}{\left(1\right)}^2+20\left(1\right)\right)-\left(\frac{1}{4}{\left(0\right)}^4-\frac{2}{3}{\left(0\right)}^3-\frac{19}{2}{\left(0\right)}^2+20\left(0\right)\right)\\ & =& \left(\frac{1}{4}-\frac{2}{3}-\frac{19}{2}+20\right)-\left(0-0-0+0\right)\\ & =& \frac{121}{12}-0\\ & =& \frac{121}{12}\end{array}$

$R_2$ is the region between $x=1$ and $x=3$

It lies totally below the $x$-axis, so the integral will give us the negative version of the area

$\begin{array}{rcl}-\left(\mathrm{Area}\mathrm{of}{R}_2\right)& =& {\int }_1^3\left(x^3-2x^2-19x+20\right)\mathrm{d}x\\ & =& {\left[\frac{1}{4}{x}^4-\frac{2}{3}{x}^3-\frac{19}{2}{x}^2+20x\right]}_1^3\\ & =& \left(\frac{1}{4}{\left(3\right)}^4-\frac{2}{3}{\left(3\right)}^3-\frac{19}{2}{\left(3\right)}^2+20\left(3\right)\right)-\left(\frac{1}{4}{\left(1\right)}^4-\frac{2}{3}{\left(1\right)}^3-\frac{19}{2}{\left(1\right)}^2+20\left(1\right)\right)-\\ & =& \left(\frac{81}{4}-18-\frac{171}{2}+60\right)-\left(\frac{1}{4}-\frac{2}{3}-\frac{19}{2}+20\right)\\ & =& -\frac{93}{4}-\frac{121}{12}\\ & =& -\frac{400}{12}\end{array}$

So the area of $R_2$ is $\frac{400}{12}$ $\left(=\frac{100}{3}\right)$

Add the two areas together to get the total area

$\mathrm{total}\mathrm{area}=\frac{121}{12}+\frac{400}{12}=\frac{521}{12}$

$\frac{521}{12}{\mathrm{units}}^2$