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: 

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

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.

​$$A_1$$ 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.