Areas Between Curves (Edexcel IGCSE Further Pure Maths): Revision Note

• Add information to any diagram provided

  • intersections between lines and curves

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

• If no diagram is provided, sketch one!

The region$R$ is bounded by the curve with equation $y=10x-x^2-16$ and the line with equation$y=8-x$.

(a) Sketch the graphs of the curve and the line on the same set of axes.

Be sure to Identify and label the region$R$ on your sketch, and indicate the points of intersection between the curve and the line.  You may assume without proof that the curve's $x$-axis intercepts all lie on the positive $x$-axis.

Because of the minus sign in front of $x^2$, the curve will be an 'upside down u-shaped' parabola

We need to find the points of intersection of the curve and line
Set their equations equal and solve to find the $x$-coordinates

$\begin{array}{rcl}8-x& =& 10x-x^2-16\\ x^2-11x+24& =& 0\\ \left(x-3\right)\left(x-8\right)& =& 0\end{array}$

$x=3\mathrm{or}x=8$

So the curve and line intersect when $x=3$ and when $x=8$

Substitute into the equation of the line to find the corresponding $y$-coordinates

$\begin{array}{l}x=3:y=8-3=5\\ \\ x=8:y=8-8=0\end{array}$

So the points of intersection are $\left(3,5\right)$ and $\left(8,0\right)$

That gives us enough information to sketch the graphs

(b) Find the area of region $R$

Here we're going to need a difference of areas:  (area under curve)$-$(area under line)

The area under the line is a right-angled triangle with vertices at $\left(3,0\right)$, $\left(3,5\right)$ and $\left(8,0\right)$

So the height is $5-0=5$ and the base is $8-3=5$

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times 5\times 5=\frac{25}{2}$

For the area under a curve we need to use a definite integral between $x=3$ and $x=8$

$\begin{array}{rcl}\mathrm{Area}\mathrm{under}\mathrm{the}\mathrm{curve}& =& {\int }_3^8\left(10x-x^2-16\right)\mathrm{d}x\\ & =& {\left[5x^2-\frac{1}{3}{x}^3-16x\right]}_3^8\\ & =& \left(5{\left(8\right)}^2-\frac{1}{3}{\left(8\right)}^3-16\left(8\right)\right)-\left(5{\left(3\right)}^2-\frac{1}{3}{\left(3\right)}^3-16\left(3\right)\right)\\ & =& \left(320-\frac{512}{3}-128\right)-\left(45-9-48\right)\\ & =& \frac{64}{3}-\left(-12\right)\\ & =& \frac{100}{3}\end{array}$

For the area of $R$, subtract the area of the triangle from the area under the curve

$\mathrm{Area}\mathrm{of}\mathrm{region}R=\frac{100}{3}-\frac{25}{2}=\frac{125}{6}$

$\frac{\mathbf{125}}{\mathbf{6}}{\mathbf{units}}^{\mathbf{2}}$

• Add information to any given diagram as you work through a question

  • intersections between curves

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

• If no diagram is provided sketch one

The diagram below shows the curves with equations $y=\mathrm{f}\left(x\right)$ and $y=\mathrm{g}\left(x\right)$, where  $\mathrm{f}\left(x\right)=(x-2){(x-3)}^2$  and  $\mathrm{g}\left(x\right)=x^2-5x+6$.

Find the area of the shaded region.

Start by finding the points of intersection

Set the equations of the curves equal to each other, and solve to find the $x$-coordinates

$\left(x-2\right){\left(x-3\right)}^2=x^2-5x+6$

It's tempting to expand the brackets on the left-hand side of the equation
Actually. here it will be more useful to factorise the right-hand side

$\begin{array}{rcl}\left(x-2\right){\left(x-3\right)}^2& =& \left(x-2\right)\left(x-3\right)\\ \left(x-2\right){\left(x-3\right)}^2-\left(x-2\right)\left(x-3\right)& =& 0\\ \left(x-2\right)\left(x-3\right)\left(\left(x-3\right)-1\right)& =& 0\\ \left(x-2\right)\left(x-3\right)\left(x-4\right)& =& 0\end{array}$

$x=2,3\mathrm{or}4$

Note that we don't need to know the corresponding $y$-coordinates here!

We have two regions here

The first region is from $x=2$ to $x=3$, and $\mathrm{f}\left(x\right)$ is the 'upper curve'

$\begin{array}{rcl}\mathrm{Region}1& =& {\int }_2^3\left(\left(x-2\right){\left(x-3\right)}^2-\left(x^2-5x+6\right)\right)\mathrm{d}x\\ & =& {\int }_2^3\left(x^3-9x^2+26x-24\right)\mathrm{d}x\\ & =& {\left[\frac{1}{4}{x}^4-3x^3+13x^2-24x\right]}_2^3\\ & =& \left(\frac{1}{4}{\left(3\right)}^4-3{\left(3\right)}^3+13{\left(3\right)}^2-24\left(3\right)\right)-\left(\frac{1}{4}{\left(2\right)}^4-3{\left(2\right)}^3+13{\left(2\right)}^2-24\left(2\right)\right)\\ & =& \left(\frac{81}{4}-81+117-72\right)-\left(4-24+52-48\right)\\ & =& -\frac{63}{4}-\left(-16\right)\\ & =& \frac{1}{4}\end{array}$

The second region is from $x=3$ to $x=4$ and $\mathrm{g}\left(x\right)$ is the 'upper curve'

$\begin{array}{rcl}\mathrm{Region}2& =& {\int }_3^4\left(\left(x^2-5x+6\right)-\left(x-2\right){\left(x-3\right)}^2\right)\mathrm{d}x\\ & =& {\int }_3^4\left(-x^3+9x^2-26x+24\right)\mathrm{d}x\\ & =& {\left[-\frac{1}{4}{x}^4+3x^3-13x^2+24x\right]}_3^4\\ & =& \left(-\frac{1}{4}{\left(4\right)}^4+3{\left(4\right)}^3-13{\left(4\right)}^2+24\left(4\right)\right)-\left(-\frac{1}{4}{\left(3\right)}^4+3{\left(3\right)}^3-13{\left(3\right)}^2+24\left(3\right)\right)\\ & =& \left(-64+192-208+96\right)-\left(-\frac{81}{4}+81-117+72\right)\\ & =& 16-\frac{63}{4}\\ & =& \frac{1}{4}\end{array}$

Now just add the two areas together to get the total area

$\mathrm{Total}\mathrm{area}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Area of the shaded region $\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{units}}^{\mathbf{2}}$