Recall that the area under a curve and above thex-axis can be computed by the definite integral. If we have two curves
( y = f(x) ) and ( y=g(x) )
By integrating the difference of two functions, you can find the area between them. Created by Sal Khan.Practice this lesson yourself on KhanAcademy.org righ.
such that
[ f(x) > g(x) nonumber]
then the area between them bounded by the horizontal lines (x = a) and (x = b) is
[ text{Area}=int_{c}^{b} left [ f(x) - g(x) right ] ;dx. nonumber]
To remember this formula we write
[ text{Area}=int_{a}^{b}text{(Top-Bottom)};dx nonumber]
Example 1
Find the area between the curves ( y=x^2) and (y=x^3).
Solution
First we note that the curves intersect at the points ((0,0)) and ((1,1)). Then we see that
[ x^3 < x^2 nonumber]
in this interval. Hence the area is given by
[begin{align*} int_{0}^{1} left( x^2 - x^3 right) dx &= {left[ frac{1}{3}x^3 - frac{1}{4}x^4 right]}_0^1 &= dfrac{1}{3} - dfrac{1}{4} &= dfrac{1}{12}. end{align*}]
Example 2
Find the area between the curves ( x = 1 - y^2 ) and ( x = y^2-1 ).
Solution
Here the curves bound the region from the left and the right.
We use the formula
[ text{Area}=int_{c}^{b}text{(Right-Left)};dy. nonumber]
For our example:
[begin{align*} int_{-1}^{1}big[ (1-y^2)-(y^2-1) big] dy &= int_{-1}^{1}(2-y^2) dy &= left(2y-dfrac{2}{3}y^3right]_{-1}^1 &=big(2-dfrac{2}{3}big)-big(-2-dfrac{2}{3} big) &= dfrac{8}{3}. end{align*}]
Example 3
Find the area between the curves ( y =0 ) and (y = 3 left( x^3-x right) ).
Solution
When we graph the region, we see that the curves cross each other so that the top and bottom switch. Hence we split the integral into two integrals:
[begin{align*} int_{-1}^{0}big[ 3(x^3-x)-0big] dx +int_{0}^{1}big[0-3(x^3-x) big] dx &= left(dfrac{3}{4}x^4-dfrac{3x^2}{2}right]_{-1}^0 - left(dfrac{3}{4}x^4-dfrac{3x^2}{2}right]_0^1 &=big(-dfrac{3}{4}+dfrac{3}{2} big) - big(dfrac{3}{4}-dfrac{3}{2} big) &=dfrac{3}{2} end{align*}.]
Let (y = f(x)) be the demand function for a product and (y = g(x)) be the supply function. Then we define the equilibrium point to be the intersection of the two curves. The consumer surplus is defined by the area above the equilibrium value and below the demand curve, while the producer surplus is defined by the area below the equilibrium value and above the supply curve.
Example 4
Find the producer surplus for the demand curve
[ f(x) = 1000 - 0.4x^2 nonumber]
and the supply curve of
[ g(x) = 42x. nonumber]
Solution
We first find the equilibrium point:
We set
[ 1000 - 0.4x^2 = 42x nonumber ]
or
[ 0.4x^2 + 42x - 1000 = 0. nonumber]
We get
[x=20 nonumber]
hence
[y=42(20)=840. nonumber]
We integrate
[ begin{align*} int_{0}^{20} left ( 840 - 42x right ) dx &= {left[ 840x-21x^2 right] }_0^{20} [4pt] &= 8400. end{align*}]
Exercises
Integrated by Justin Marshall.
Area between Curves
The area between curves is given by the formulas below.
Area = (int_a^b {,left {fleft( x right) - gleft( x right)} right ,dx} ) | |
for a region bounded above by | |
for a region bounded on the left by | |
Example 1:^{1} | |
(eqalign{{rm{Area}} &= int_0^1 {left {x - {x^2}} right dx} &= int_0^1 {left( {x - {x^2}} right)dx} &= left. {left( {frac{1}{2}{x^2} - frac{1}{3}{x^3}} right)} right _0^1 &= left( {frac{1}{2} - frac{1}{3}} right) - left( {0 - 0} right) &= frac{1}{6}}) | |
1 | Find the area between |
(eqalign{{rm{Area}} &= int_{ - 1}^1 {left {y + 3 - {y^2}} right dy} &= int_{ - 1}^1 {left( {y + 3 - {y^2}} right)dy} &= left. {left( {frac{1}{2}{y^2} + 3y - frac{1}{3}{x^3}} right)} right _{ - 1}^1 &= left( {frac{1}{2} + 3 - frac{1}{3}} right) - left( {frac{1}{2} - 3 + frac{1}{3}} right) &= frac{{16}}{3}}) |
See also
Area under a curve, definite integral, absolute value rules