Finding areas using integration

​​In a nutshell

It is possible to use the new integration techniques you have learnt to find the area between the curves of many different functions and the $$x$$-axis. Integration can also be used to find the area between two curves.​

Area between two curves

The area bounded between the curves of two functions: $$y=f(x)$$ and $$y=g(x)$$ in an interval $$a\leq x \leq b$$ is given to be:  

$$\begin{aligned}A&= \text{area under} \ f(x) - \text{area under} \ g(x)\\&=\int_a^b f(x)\, dx - \int_a^b g(x)\,dx\end{aligned} \\ \ \\ \boxed{A=\int_a^b (f(x) - g(x))\,dx}$$​​

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Note: In the case where the two functions intersect in the interval $$a \lt x \lt b$$, this formula does not work. You need to break up the region into multiple areas and apply this formula to each area separately, taking care to keep the areas positive.

Example 1

The graph below shows the curve of $$y=\dfrac{3+\sin(x)}{\cos^2(x)}$$ in the domain $$-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$$. Find the area of the shaded region.

Use definite integration and trigonometric identities to find the area:

​$$\begin{aligned} A&=\int_0^\frac{\pi}{4} \dfrac{3+\sin(x)}{\cos^2(x)}\,dx\\&= \int_0^\frac{\pi}{4} \left( \dfrac{3}{\cos^2(x)} + \dfrac{\sin(x)}{\cos^2(x)} \right) \, dx \\&=\int_0^\frac{\pi}{4} \left( 3\sec^2(x) + \sec(x)\tan(x)\right)\,dx\\&=\left[3\tan(x) + \sec(x)\right]_0^\frac{\pi}{4}\,dx \\&=\left(3\tan\left(\dfrac{\pi}{4}\right) + \sec\left(\dfrac{\pi}{4}\right)\right)-\left(3\tan(0) + \sec(0)\right) \\&=\left(3(1)+(\sqrt2)\right) - \left(3(0)+(1)\right)\\&=2+\sqrt2 \end{aligned}$$​​

​$$\int_0^\frac{\pi}{4} \dfrac{3+\sin(x)}{\cos^2(x)}\,dx=\underline{2+\sqrt{2}}$$​​

Example 2

Find the area of the shaded region below:

First find the points of intersection:

​$$\begin{aligned}x^2e^x &= xe^x \\ x^2&=x\\x&=0,1 \end{aligned}$$​​

Find an expression for the desired area:

​$$\begin{aligned} A&=\int_0^1 xe^x \, dx - \int_0^1 x^2 e^x \,dx\\&=\int_0^1 (x-x^2)e^x\,dx \end{aligned}$$​​

Compute the integral using integration by parts twice:

​$$\begin{aligned}u=x-x^2 &\rightarrow \dfrac{du}{dx}=1-2x\\ \dfrac{dv}{dx}=e^x &\rightarrow v=e^x \end{aligned}$$​​

​$$\begin{aligned} A&=\int_0^1 (x-x^2)e^x\,dx\\&=\left[uv\right]_0^1 - \int_0^1 v \dfrac{du}{dx}\,dx\\&=\left[(x-x^2)e^x\right]_0^1 - \int_0^1 (e^x)(1-2x)\,dx\\&=\left( (0) - (0)\right) +\int_0^1 (2x-1)e^x\,dx\\&=\int_0^1 (2x-1)e^x\,dx \end{aligned}$$​​

​$$\begin{aligned} u=2x-1 &\rightarrow \dfrac{du}{dx}=2\\ \dfrac{dv}{dx}=e^x &\rightarrow v=e^x \end{aligned}$$​​

​$$\begin{aligned} \int_0^1 (2x-1)e^x\,dx &= \left[(2x-1)e^x\right]_0^1 - \int_0^1 (e^x)(2)\,dx\\&=\left( (1e^1) - (-1e^0)\right) - \int_0^1 2e^x\,dx\\&=e +1 - \left[2e^x\right]_0^1\\&=e+1- \left(2e - 2\right)\\&=3-e \end{aligned}$$​​

​$$A= \underline{3-e}$$​​

​Note: The order of subtraction is important. Always subtract the "lower" area from the "upper" area.