Area Between Curves

Video lesson · from 32:533 worked examples

One lesson video covers all of Integration, so it opens at 32:53 for this topic — not from the beginning.

Theory

In general, we can find the area enclosed between two curves from $x=a$ to $x=b$ by using:

\int_a^b \big( f(x) - g(x) \big) \,dx

where $f(x)$ is the upper curve and $g(x)$ is the lower curve.

⚠️ Common Examiner Traps

Brackets round the second curve: bracket errors are the most common way to lose marks on this question. Write $\big(f(x) - g(x)\big)$ — subtracting $g(x)$ means subtracting every term of it, signs included.
Upper minus lower, not left minus right: decide which curve is on top over the interval. Getting it the wrong way round gives the right size with the wrong sign.
Find the limits if you are not given them: set the two curves equal and solve. Do not assume the limits are the axis intercepts.
Split at every crossing: if the curves swap over inside the interval, the integral must be split there and each piece taken as upper minus lower separately.
Calculator discipline: evaluate carefully and keep your lines of working consistent.

Worked examples

Example 1

Calculate the shaded area enclosed by $y = 4x + 16$ and $y = 2x^2 + 10$ .

Integrate Upper curve - Lower curve between the intersection points $x=-1$ and $x=3$ :

\begin{aligned} A &= \int_{-1}^3 \big( (4x+16) - (2x^2+10) \big) \,dx \\ &= \int_{-1}^3 ( -2x^2 + 4x + 6 ) \,dx \\ &= \left[ -\frac{2x^3}{3} + 2x^2 + 6x \right]_{-1}^3 \\ &= \left( -18 + 18 + 18 \right) - \left( \frac{2}{3} + 2 - 6 \right) \\ &= 18 - \left( -\frac{10}{3} \right) \\ &= \frac{54}{3} + \frac{10}{3} = \frac{64}{3} \text{ square units} \end{aligned}

Example 2

Calculate the shaded area enclosed by the curves with equations $y = 6 - 3x^2$ and $y = -3 - 2x^2$ .

First, find the points of intersection:

\begin{aligned} 6 - 3x^2 &= -3 - 2x^2 \\ x^2 &= 9 \\ x &= 3, x &= -3 \end{aligned}

The limits are $-3$ and $3$ .

\begin{aligned} A &= \int_{-3}^3 \big( (6-3x^2) - (-3-2x^2) \big) \,dx \\ &= \int_{-3}^3 ( 9 - x^2 ) \,dx \\ &= \left[ 9x - \frac{x^3}{3} \right]_{-3}^3 \\ &= \left( 27 - 9 \right) - \left( -27 + 9 \right) \\ &= 18 - (-18) = 36 \text{ square units} \end{aligned}

Example 3

Two functions are defined by $x \in \mathbb{R}$ by $f(x) = x^3 - 7x^2 + 8x + 16$ and $g(x) = 4x + 4$ .

Calculate the shaded area.

The curves intersect at multiple points. $A_1$ is from $x = -1$ to $x = 2$ where $f(x)$ is upper. $A_2$ is from $x = 2$ to $x = 6$ where $g(x)$ is upper.

Find $A_1$ :

\begin{aligned} A_1 &= \int_{-1}^2 \big( f(x) - g(x) \big) \,dx \\ &= \int_{-1}^2 (x^3 - 7x^2 + 4x + 12) \,dx \\ &= \left[ \frac{x^4}{4} - \frac{7x^3}{3} + 2x^2 + 12x \right]_{-1}^2 \\ &= \left( 4 - \frac{56}{3} + 8 + 24 \right) - \left( \frac{1}{4} + \frac{7}{3} + 2 - 12 \right) \\ &= \left( \frac{52}{3} \right) - \left( -\frac{89}{12} \right) = \frac{208 + 89}{12} = \frac{297}{12} = \frac{99}{4} \end{aligned}

Find $A_2$ :

\begin{aligned} A_2 &= \int_2^6 \big( g(x) - f(x) \big) \,dx \\ &= \int_2^6 (-x^3 + 7x^2 - 4x - 12) \,dx \\ &= \left[ -\frac{x^4}{4} + \frac{7x^3}{3} - 2x^2 - 12x \right]_2^6 \\ &= \left( -324 + 504 - 72 - 72 \right) - \left( -4 + \frac{56}{3} - 8 - 24 \right) \\ &= ( 36 ) - \left( -\frac{52}{3} \right) = \frac{160}{3} \end{aligned}

Total Area = $\frac{99}{4} + \frac{160}{3} = \frac{297 + 640}{12} = \frac{937}{12} \approx 78.08$