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Area under the x-axis

Area under the x-axis

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OCR A

OCR AAQAPearson Edexcel

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Tutor: Bilal

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 xx-axis. However, calculating area becomes more complicated when asked to consider functions that have graphs which are below the xx​-axis.



Negative integrals

If the graph is below the xx-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), where f(x)=x2+10x+16f(x)=x^2+10x+16. What is the area of the graph that is bounded between the lines x=8x=-8 and x=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 xx-axis in the region 2x8-2\leq x \leq -8. Therefore, the integral will be negative. Computing the integral:

82(x2+10x+16)dx=[13x3+5x2+16x]82=(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}​​


Remember that the area is positive:

The area is 36\underline{36}.



Integrals made up of positive and negative sections

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



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


The graph shows that f(x)f(x) is negative for 4x<2-4\leq x\lt -2, and positive for 2<x0-2\lt x \leq 0. Therefore, compute these two integrals separately.

42(x2+10x+16)dx=[13x3+5x2+16x]42=(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}​​


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


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}


The area for 2<x0-2\lt x \leq 0 is 443\dfrac{44}{3}.


Add the two separate areas to find the total area:

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


The total area is 24\underline{24}.


 


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Learn with Basics

Length:
Indefinite integrals

Unit 1

Indefinite integrals

Area under a curve

Unit 2

Area under a curve

Jump Ahead

Optional

Area under the x-axis

Unit 3

Area under the x-axis

Final Test

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FAQs - Frequently Asked Questions

How do you work out an area bound by a curve and 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?

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?

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