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Wize AP Calculus (AB) Textbook > Applications of Integration

Area Between Curves

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Area Between Curves

Definite integrals give us the exact area under the curve of a graph. What about if we wanted to find the area between curves? The integral of the difference of two functions answers this question.



Area Between Two Curves

If f(x)≥g(x)f(x)\geq g(x) for xx in [a,b][a,b], then the area above g(x)g(x) and below f(x)f(x) between x=ax=a and x=bx=b is given by

∫ab[f(x)−g(x)]dx\boxed{\int_a^b\left[f(x)-g(x)\right]dx}


Watch Out!
If the functions intersect multiple times, it may result in multiple regions that are bounded between the two curves.


We need to find the point of intersection between the curves, then we may need to break up the integral into multiple parts to ensure each area added up is positive.

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Example: Area Between Curves

Find the area enclosed by y=2x2+10y=2x^2+10 and y=4x+16y=4x+16 between x=0x=0 and x=5x=5.




Intersection
2x2+10=4x+162x2−4x−6=0(2x−6)(x+1)=0x=3,−1\begin{aligned} &2x^2+10=4x+16\\ &2x^2-4x-6=0\\ &(2x-6)(x+1)=0\\ &x=3,-1 \end{aligned}

Area Between The Curves

∫03[(4x+16)−(2x2+10)]dx+∫35[(2x2+10)−(4x+16)]dx\displaystyle \int_0^3[(4x+16)-(2x^2+10)]dx+\int_3^5[(2x^2+10)-(4x+16)]dx

∫03(−2x2+4x+6)dx+∫35(2x2−4x−6)dx\displaystyle \int_0^3(-2x^2+4x+6)dx+\int_3^5(2x^2-4x-6)dx

=(−2x33+2x2+6x)∣03+(2x33−2x2−6x)∣35\displaystyle =\left(\frac{-2x^3}{3}+2x^2+6x\right)\bigg|_0^3+\left(\frac{2x^3}{3}-2x^2-6x\right)\bigg|_3^5

=(−2(3)33+2(3)2+6(3))−(−2(0)33+2(0)2+6(0))+(2(5)33−2(5)2−6(5))−(2(3)33−2(3)2−6(3))\displaystyle =\left(\frac{-2(3)^3}{3}+2(3)^2+6(3)\right)\displaystyle -\left(\frac{-2(0)^3}{3}+2(0)^2+6(0)\right)+\left(\frac{2(5)^3}{3}-2(5)^2-6(5)\right)-\left(\frac{2(3)^3}{3}-2(3)^2-6(3)\right)

=(−18+18+18)+(2503−50−30)−(18−18−18)\displaystyle =\left(-18+18+18\right)\displaystyle +\left(\frac{250}{3}-50-30\right)-\left(18-18-18\right)

=18+(2503−80)+18\displaystyle =18+\left( \frac{250}{3}-80\right)+18

=36+(2503−2403)\displaystyle = 36+\left(\frac{250}{3}-\frac{240}{3}\right)

=36+103≈39.33=\displaystyle 36+\frac{10}{3} \approx39.33
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Example: Area Between Curves

Find the area of the shaded region below


Rearranging for x: x=2−y2  and  x=2−yx=2-y^2\ \text{ and }\ x=2-y

Point of Intersection:
2−y2=2−y2-y^2=2-y
0=2−y−2+y20=2-y-2+y^2
0=y2−y0=y^2-y
0=y(y−1)0=y\left(y-1\right)
y=0,1y=0,1

Based on the graph, x=2−y2x=2-y^2 is the rightmost curve and x=2−yx=2-yis the leftmost curve.

A=∫01[(2−y2)−(2−y)]dy\displaystyle A=\int_0^1\left[\left(2-y^2\right)-\left(2-y\right)\right]dy

=∫01[2−y2−2+y]dy\displaystyle =\int_0^1\left[2-y^2-2+y\right]dy

=∫01[−y2+y]dy\displaystyle =\int_0^1\left[-y^2+y\right]dy

=[−y33+y22]01\displaystyle =\left[-\frac{y^3}{3}+\frac{y^2}{2}\right]_0^1

=(−13+12)−0\displaystyle =\left(-\frac{1}{3}+\frac{1}{2}\right)-0

=−26+36=16\displaystyle =-\frac{2}{6}+\frac{3}{6}=\boxed{\frac{1}{6}}

Practice: Area Between Curves

Find the area bounded by the curves y=x2−x  and  y=3x−3 y=x^2-x\ \text{ and }\ y=3x-3\