Area under the x-axis

In a nutshell

Integration can be used to find the area between the graph of a function and the $x$ -axis. However, calculating area becomes more complicated when asked to consider functions that have graphs which are below the $x$ -axis.

Negative integrals

If the graph is below the $x$ -axis, then the integral is negative. Although an integral can be negative, it is important to remember that area is defined to be a positive quantity.

Example 1

Below is the graph of the function $y=f(x)$ , where $f(x)=x^2+10x+16$ . What is the area of the graph that is bounded between the lines $x=-8$ and $x=-2$ ?

Maths; Integration I; KS5 Year 12; Area under the x-axis

The sketch of the function shows that the graph is entirely below the $x$ -axis in the region $-2\leq x \leq -8$ . Therefore, the integral will be negative. Computing the integral:

$\begin{aligned}\int_{-8}^{-2}(x^2+10x+16)dx&=\left[\dfrac{1}{3}x^3 +5x^2 +16x\right]_{-8}^{-2}\\&=\left(\dfrac{1}{3}(-2)^3+5(-2)^2+16(-2)\right)-\left(\dfrac{1}{3}(-8)^3+5(-8)^2+16(-8)\right)\\&=\left(-\dfrac{44}{3}\right)-\left(\dfrac{64}{3}\right)\\&=-36\end{aligned}$

Remember that the area is positive:

The area is $\underline{36}$ .

Integrals made up of positive and negative sections

When an integral of a function $f(x)$ has limits such that the graph of $y=f(x)$ between these limits has some parts above the $x$ -axis and some parts below the $x$ -axis, you need to split up the integral into different parts depending on where the function is positive or negative.

Example 2

Find the area of the region bounded between the graph of $y=f(x)$ (where $f(x)=x^2+10x+16$ ), the $x$ -axis and the lines $x=0$ and $x=-4$ .

Maths; Integration I; KS5 Year 12; Area under the x-axis

The graph shows that $f(x)$ is negative for $-4\leq x\lt -2$ , and positive for $-2\lt x \leq 0$ . Therefore, compute these two integrals separately.

$\begin{aligned}\int_{-4}^{-2}(x^2+10x+16)dx&=\left[\dfrac{1}{3}x^3 +5x^2 +16x\right]_{-4}^{-2}\\&=\left(\dfrac{1}{3}(-2)^3+5(-2)^2+16(-2)\right)-\left(\dfrac{1}{3}(-4)^3+5(-4)^2+16(-4)\right)\\&=\left(-\dfrac{44}{3}\right)-\left(-\dfrac{16}{3}\right)\\&=-\dfrac{28}{3}\end{aligned}$

The area for $-4\leq x\lt -2$ is $\dfrac{28}{3}$ .

$\begin{aligned}\int_{-2}^{0}(x^2+10x+16)dx&=\left[\dfrac{1}{3}x^3 +5x^2 +16x\right]_{-2}^{0}\\&=\left(\dfrac{1}{3}(0)^3+5(0)^2+16(0)\right)-\left(\dfrac{1}{3}(-2)^3+5(-2)^2+16(-2)\right)\\&=\left(0\right)-\left(-\dfrac{44}{3}\right)\\&=\dfrac{44}{3}\end{aligned}$

The area for $-2\lt x \leq 0$ is $\dfrac{44}{3}$ .

Add the two separate areas to find the total area:

$\dfrac{28}{3}+\dfrac{44}{3}=\dfrac{72}{3}=24$

The total area is $\underline{24}$ .