Area between a curve and a line
In a nutshell
There are two methods to find the area between a curve and a line. One method is to break up the area into the sum or difference of an integral and a polygon, and the other method is about combining two integrals into one.
Breaking up areas
One method for finding the area between a curve and a line is to break up the desired region into the area between a curve and the x-axis (which can be found via definite integration) and the area of a polygon. The polygon is often either a rectangle, triangle or trapezium. The desired area will be the sum or difference of these two areas.
Example 1
Find the area (A) of the shaded region in the diagram below. The line s is the equation y=3−x, and the curve u has equation y=x2+x.
Note that this area can be broken up into two smaller areas: A1 and A2.
A1 is the area between the curve and the x-axis from the origin up to the point where s and u intersect. A2 is a right-angled triangle.
To calculate the areas, first find the points of intersection between the line and the curve:
3−xx2+2x−3(x+3)(x−1)x=x2+x=0=0=1,−3
The point of intersection in the diagram concerns a positive x, so x=1.
Using this, calculate the area of A1 using definite integration:
A1=∫01(x2+x)dx=[31x3+21x2]01=(31(1)3+21(1)2)−(31(0)3+21(0)2)=65
A1=65
A2 is the area of a right-angled triangle. So, find the base and perpendicular height.
To find the height, substitute x=1 into the linear equation:
y=3−x=3−1=2
The perpendicular height is 2.
To find the base length, find the x-intercept of the line, and subtract 1 from this value:
yy3−xx3−1=3−x=0=0=3=2
The base length is 2.
The area of the right-angled triangle can now be calculated:
A2=21bh=21(2)(2)=2
A2=2
The region A is the sum of the two smaller areas, which can now be calculated:
A=A1+A2=65+2=617
A=617
Combining integrals
Another way to calculate the area between a curve and a line is to represent the two areas as integrals and combine them to give a single integral - this only works if the two integrals have equal limits, as it makes use of the additive property of integration:
∫ba(f(x)±g(x))dx=∫baf(x)dx±∫bag(x)dx
Example 2
Find the area of the shaded region (A) below, given that f(x)=3+x−x2 and 3−x.
First, find the values of x for which the two functions intersect:
3+x−x2x2−2xx=−x+3=0=0,2
The area of the shaded region can therefore be expressed as the difference of two integrals:
A=∫02f(x)dx−∫02g(x)dx
As the limits are the same for both integrals, the additive property can be applied, expressing the area as a single integral:
A=∫02(3+x−x2)dx−∫02(−x+3)dx=∫02((3+x−x2)−(−x+3))dx=∫02(2x−x2)dx=[x2−31x3]02=((2)2−31(2)3)−((0)2−31(0)3)=34
A=34
Note: The order of subtraction is important. Always subtract the "lower" area from the "upper" area.