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 $y=x^2+x$ .

Maths; Integration I; KS5 Year 12; Area between a curve and a line

Note that this area can be broken up into two smaller areas: $A_1$ and $A_2$ .

$A_1$ is the area between the curve and the $x$ -axis from the origin up to the point where $s$ and $u$ intersect. $A_2$ is a right-angled triangle.

Maths; Integration I; KS5 Year 12; Area between a curve and a line

To calculate the areas, first find the points of intersection between the line and the curve:

$\begin{aligned} 3-x&=x^2+x\\x^2+2x-3&=0\\(x+3)(x-1)&=0\\x&=1,-3 \end{aligned}$

The point of intersection in the diagram concerns a positive $x$ , so $x=1$ .

Using this, calculate the area of $A_1$ using definite integration:

$\begin{aligned} A_1&= \int_0^1 (x^2+x)\, dx\\&=\left[\dfrac13 x^3 + \dfrac12 x^2\right]_0^1\\&=\left(\dfrac13 (1)^3+\dfrac12 (1)^2\right) - \left(\dfrac13 (0)^3 +\dfrac12 (0)^2\right)\\&=\dfrac56 \end{aligned}$

$A_1=\dfrac56$

$A_2$ 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:

$\begin{aligned}y&=3-x\\&=3-1\\&=2 \end{aligned}$

The perpendicular height is $2$ .

To find the base length, find the $x$ -intercept of the line, and subtract $1$ from this value:

$\begin{aligned} y&=3-x\\y&=0\\3-x&=0\\x&=3\\3-1&=2 \end{aligned}$

The base length is $2$ .

The area of the right-angled triangle can now be calculated:

$\begin{aligned} A_2&=\dfrac12 bh\\&=\dfrac12(2)(2)\\&=2 \end{aligned}$

$A_2=2$

The region $A$ is the sum of the two smaller areas, which can now be calculated:

$\begin{aligned} A&=A_1+A_2\\&=\dfrac56+2\\&=\dfrac{17}{6} \end{aligned}$

$\underline{A=\dfrac{17}{6}}$

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:

$\int_b^a(f(x)\pm g(x))\, dx= \int_b^a f(x)\, dx \pm \int_b^a g(x) \, dx$

Example 2

Find the area of the shaded region ( $A$ ) below, given that $f(x)=3+x-x^2$ and $3-x$ .

Maths; Integration I; KS5 Year 12; Area between a curve and a line

First, find the values of $x$ for which the two functions intersect:

$\begin{aligned} 3+x-x^2&=-x+3\\x^2-2x&=0\\x&=0,2 \end{aligned}$

The area of the shaded region can therefore be expressed as the difference of two integrals:

$A=\int_0^2f(x)\, dx - \int_0^2 g(x)\, dx$ 

As the limits are the same for both integrals, the additive property can be applied, expressing the area as a single integral:

$\begin{aligned}A&= \int_0^2(3+x-x^2)\, dx - \int_0^2(-x+3)\, dx\\&=\int_0^2 \left((3+x-x^2)-(-x+3)\right)\,dx\\&=\int_0^2(2x-x^2)\,dx\\&=\left[x^2-\dfrac13 x^3\right]_0^2\\&=\left((2)^2 - \dfrac13 (2)^3\right) - \left((0)^2 - \dfrac13 (0)^3\right)\\&=\dfrac43 \end{aligned}$

$\underline{A=\dfrac43}$ 

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