Area under a curve

In a nutshell

Definite integration can be used to calculate the area between the graph of a function $y=f(x)$ and the $x$ -axis between two limits, $x=a$ and $x=b$ .

Integration and area

The area ( $A$ ) of the region bounded between the graph of $y=f(x)$ , the $x$ -axis and the lines $x=a,x=b$ is given by:

$\begin{aligned}A&=\int_b^a f(x)\,dx\\&=\lim_{\delta x \rightarrow 0}\sum_{x=b}^af(x)\delta x\end{aligned}$

Maths; Integration I; KS5 Year 12; Area under a curve

Note: You don't need to know what $\lim_{\delta x \rightarrow 0}\sum_{x=b}^af(x)\delta x$ means, you just need to know that this can be used as another notation to denote the area represented by an integral.

Example 1

In the diagram below, find the values of $A_1$ and $A_2$ using integration.

Maths; Integration I; KS5 Year 12; Area under a curve

$A_1$ is the area between $f(x)=-x$ and the $x$ -axis, bounded by $x=0$ and $x=-4$ . The area is therefore given by:

$\begin{aligned}A_1&= \int_{-4}^0-x \,dx\\&=\left[-\dfrac{1}{2}x^2\right]_{-4}^0\\&=\left(-\dfrac{1}{2}(0)^2\right)-\left(-\dfrac{1}{2}(-4)^2\right)\\&=-(-8)\\&=8\end{aligned}$

$\underline{A_1=8}$

$A_2$ is the area between $f(x)=\sqrt{x}$ and the $x$ -axis, bounded by $x=4$ and $x=0$ . The area is therefore given by:

$\begin{aligned}A_2&=\int_0^4\sqrt{x}\,dx\\&=\left[\dfrac{1}{\frac{3}{2}} x^{\frac{3}{2}}\right]_0^4\\&=\left[\dfrac{2}{3} x^{\frac{3}{2}}\right]_0^4\\&=\left(\dfrac{2}{3}(4)^\frac{3}{2}\right)-\left(\dfrac{2}{3}(0)^\frac{3}{2}\right)\\&=\dfrac{16}{3}\end{aligned}$

$\underline{A_2=\dfrac{16}{3}}$

Note: When finding areas on graphs, there is no need for units unless the question explicitly states them.