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Area under the x-axis

Area under the x-axis

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Summary

Area under the x-axis

In a nutshell

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



Negative integrals

If the graph is below the xxx-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)y=f(x)y=f(x), where f(x)=x2+10x+16f(x)=x^2+10x+16f(x)=x2+10x+16. What is the area of the graph that is bounded between the lines x=−8x=-8x=−8 and x=−2x=-2x=−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 xxx-axis in the region −2≤x≤−8-2\leq x \leq -8−2≤x≤−8. Therefore, the integral will be negative. Computing the integral:

​∫−8−2(x2+10x+16)dx=[13x3+5x2+16x]−8−2=(13(−2)3+5(−2)2+16(−2))−(13(−8)3+5(−8)2+16(−8))=(−443)−(643)=−36\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}∫−8−2​(x2+10x+16)dx​=[31​x3+5x2+16x]−8−2​=(31​(−2)3+5(−2)2+16(−2))−(31​(−8)3+5(−8)2+16(−8))=(−344​)−(364​)=−36​​​


Remember that the area is positive:

The area is 36‾\underline{36}36​.



Integrals made up of positive and negative sections

When an integral of a function f(x)f(x)f(x)​ has limits such that the graph of y=f(x)y=f(x)y=f(x)​ between these limits has some parts above the xxx​-axis and some parts below the xxx​-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)y=f(x)y=f(x) (where f(x)=x2+10x+16f(x)=x^2+10x+16f(x)=x2+10x+16), the xxx-axis and the lines x=0x=0x=0 and x=−4x=-4x=−4.



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

​

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

​∫−4−2(x2+10x+16)dx=[13x3+5x2+16x]−4−2=(13(−2)3+5(−2)2+16(−2))−(13(−4)3+5(−4)2+16(−4))=(−443)−(−163)=−283\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}∫−4−2​(x2+10x+16)dx​=[31​x3+5x2+16x]−4−2​=(31​(−2)3+5(−2)2+16(−2))−(31​(−4)3+5(−4)2+16(−4))=(−344​)−(−316​)=−328​​​​


The area for −4≤x<−2-4\leq x\lt -2−4≤x<−2 is 283\dfrac{28}{3}328​.


​∫−20(x2+10x+16)dx=[13x3+5x2+16x]−20=(13(0)3+5(0)2+16(0))−(13(−2)3+5(−2)2+16(−2))=(0)−(−443)=443\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}∫−20​(x2+10x+16)dx​=[31​x3+5x2+16x]−20​=(31​(0)3+5(0)2+16(0))−(31​(−2)3+5(−2)2+16(−2))=(0)−(−344​)=344​​​


The area for −2<x≤0-2\lt x \leq 0−2<x≤0 is 443\dfrac{44}{3}344​.


Add the two separate areas to find the total area:

​283+443=723=24\dfrac{28}{3}+\dfrac{44}{3}=\dfrac{72}{3}=24328​+344​=372​=24​​


The total area is 24‾\underline{24}24​.


 


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FAQs - Frequently Asked Questions

How do you work out an area bound by a curve and the x-axis?

Integrate the curve with respect to dx. The limits of integration would be the x coordinates where the curve crosses the x-axis.

How do you work out the area between the graph of a function and the x-axis in a region where the graph has some parts below the x-axis and some parts above the x-axis?

FInd out where the graph is negative where the graph is positive. Split up the integral into these sections and compute each integral separately, making sure to take the positive of the integrals that give a "negative area".

How do you work out the area between the graph of a function and the x-axis in a region where the graph is entirely below the x-axis?

Compute the integral and turn it positive since area is positive.

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