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 $$x=-2$$?

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 $$x=-4$$.

​

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 $$\underline{24}$$.