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Finding areas using integration

Finding areas using integration

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Explainer Video

Tutor: Bilal

Summary

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 xxx-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)y=f(x)y=f(x) and y=g(x)y=g(x)y=g(x) in an interval a≤x≤ba\leq x \leq ba≤x≤b is given to be:  


A=area under f(x)−area under g(x)=∫abf(x) dx−∫abg(x) dx A=∫ab(f(x)−g(x)) dx\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}A​=area under f(x)−area under g(x)=∫ab​f(x)dx−∫ab​g(x)dx​ A=∫ab​(f(x)−g(x))dx​​​
Maths; Integration II; KS5 Year 13; Finding areas using integration

​

Note: In the case where the two functions intersect in the interval a<x<ba \lt x \lt ba<x<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=3+sin⁡(x)cos⁡2(x)y=\dfrac{3+\sin(x)}{\cos^2(x)}y=cos2(x)3+sin(x)​ in the domain −π2≤x≤π2-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}−2π​≤x≤2π​. Find the area of the shaded region.

Maths; Integration II; KS5 Year 13; Finding areas using integration


Use definite integration and trigonometric identities to find the area:

​A=∫0π43+sin⁡(x)cos⁡2(x) dx=∫0π4(3cos⁡2(x)+sin⁡(x)cos⁡2(x)) dx=∫0π4(3sec⁡2(x)+sec⁡(x)tan⁡(x)) dx=[3tan⁡(x)+sec⁡(x)]0π4 dx=(3tan⁡(π4)+sec⁡(π4))−(3tan⁡(0)+sec⁡(0))=(3(1)+(2))−(3(0)+(1))=2+2\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}A​=∫04π​​cos2(x)3+sin(x)​dx=∫04π​​(cos2(x)3​+cos2(x)sin(x)​)dx=∫04π​​(3sec2(x)+sec(x)tan(x))dx=[3tan(x)+sec(x)]04π​​dx=(3tan(4π​)+sec(4π​))−(3tan(0)+sec(0))=(3(1)+(2​))−(3(0)+(1))=2+2​​​​


​∫0π43+sin⁡(x)cos⁡2(x) dx=2+2‾\int_0^\frac{\pi}{4} \dfrac{3+\sin(x)}{\cos^2(x)}\,dx=\underline{2+\sqrt{2}}∫04π​​cos2(x)3+sin(x)​dx=2+2​​​​


Example 2

Find the area of the shaded region below:

Maths; Integration II; KS5 Year 13; Finding areas using integration


First find the points of intersection:

​x2ex=xexx2=xx=0,1\begin{aligned}x^2e^x &= xe^x \\ x^2&=x\\x&=0,1 \end{aligned}x2exx2x​=xex=x=0,1​​​


Find an expression for the desired area:

​A=∫01xex dx−∫01x2ex dx=∫01(x−x2)ex dx\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}A​=∫01​xexdx−∫01​x2exdx=∫01​(x−x2)exdx​​​


Compute the integral using integration by parts twice:

​u=x−x2→dudx=1−2xdvdx=ex→v=ex\begin{aligned}u=x-x^2 &\rightarrow \dfrac{du}{dx}=1-2x\\ \dfrac{dv}{dx}=e^x &\rightarrow v=e^x \end{aligned}u=x−x2dxdv​=ex​→dxdu​=1−2x→v=ex​​​


​A=∫01(x−x2)ex dx=[uv]01−∫01vdudx dx=[(x−x2)ex]01−∫01(ex)(1−2x) dx=((0)−(0))+∫01(2x−1)ex dx=∫01(2x−1)ex dx\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}A​=∫01​(x−x2)exdx=[uv]01​−∫01​vdxdu​dx=[(x−x2)ex]01​−∫01​(ex)(1−2x)dx=((0)−(0))+∫01​(2x−1)exdx=∫01​(2x−1)exdx​​​


​u=2x−1→dudx=2dvdx=ex→v=ex\begin{aligned} u=2x-1 &\rightarrow \dfrac{du}{dx}=2\\ \dfrac{dv}{dx}=e^x &\rightarrow v=e^x \end{aligned}u=2x−1dxdv​=ex​→dxdu​=2→v=ex​​​


​∫01(2x−1)ex dx=[(2x−1)ex]01−∫01(ex)(2) dx=((1e1)−(−1e0))−∫012ex dx=e+1−[2ex]01=e+1−(2e−2)=3−e\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}∫01​(2x−1)exdx​=[(2x−1)ex]01​−∫01​(ex)(2)dx=((1e1)−(−1e0))−∫01​2exdx=e+1−[2ex]01​=e+1−(2e−2)=3−e​​​


​A=3−e‾A= \underline{3-e}A=3−e​​​


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

 

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FAQs - Frequently Asked Questions

How do you find the area between two curves that do intersect?

Split up the region into smaller areas and find each area individually, making sure the area is positive for each section.

How do you find the area between two curves that don't intersect?

Integrate the difference of the two functions between the appropriate limits.

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