Application of Integrals • Topic 2 of 2

Area Between Two Curves

The area between two curves $y=f(x)$ (upper) and $y=g(x)$ (lower) from $x=a$ to $x=b$ is the integral of the gap between them:

$$\text{Area}=\int_a^b \big[f(x)-g(x)\big]\,dx,\qquad f(x)\ge g(x)\ \text{on }[a,b].$$

Finding the limits

The limits $a,b$ are usually the $x$-coordinates of the points of intersection, found by solving $f(x)=g(x)$. Decide which curve is on top in the region (test a sample point) so the integrand stays non-negative.

If the curves cross

When the curves swap top/bottom inside the interval, split at the crossing points and integrate each sub-interval with the correct "upper minus lower" order. The total area is the sum of these positive pieces.

Solved Examples
1
Worked Example
Find the area between $y=x$ and $y=x^2$ from their intersections.
Solution

Intersections: $x=x^2\Rightarrow x=0,1$. On $(0,1)$, $x\ge x^2$. Area $=\int_0^1 (x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac12-\dfrac13=\dfrac16$ square units.

2
Worked Example
Find the area between $y=x^2$ and $y=4$.
Solution

Intersections: $x^2=4\Rightarrow x=\pm2$. Upper curve is $y=4$. Area $=\int_{-2}^{2}(4-x^2)\,dx=\left[4x-\dfrac{x^3}{3}\right]_{-2}^{2}=\left(8-\dfrac83\right)-\left(-8+\dfrac83\right)=\dfrac{32}{3}$ square units.

3
Worked Example
Find the area between the line $y=2x$ and the parabola $y=x^2$.
Solution

$x^2=2x\Rightarrow x=0,2$. On $(0,2)$, $2x\ge x^2$. Area $=\int_0^2(2x-x^2)\,dx=\left[x^2-\dfrac{x^3}{3}\right]_0^2=4-\dfrac83=\dfrac{4}{3}$ square units.

4
Worked Example
Set up the area between $y=\sqrt{x}$ and $y=x$ for $0\le x\le1$.
Solution

On $(0,1)$, $\sqrt{x}\ge x$. Area $=\int_0^1(\sqrt{x}-x)\,dx=\left[\dfrac{2}{3}x^{3/2}-\dfrac{x^2}{2}\right]_0^1=\dfrac23-\dfrac12=\dfrac16$ square units.

Key Points

  • Area between curves $=\int_a^b(\text{upper}-\text{lower})\,dx$.
  • Limits are the intersection points: solve $f(x)=g(x)$.
  • Test a sample point to decide which curve is upper.
  • If the curves cross, split the interval and add positive pieces.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Area between $y=x$ and $y=x^2$ on $[0,1]$ is:
Explanation: $\int_0^1(x-x^2)dx=1/6$.
Q2.The limits for area between two curves come from:
Explanation: Solve $f(x)=g(x)$.
Q3.Area between $y=x^2$ and $y=4$ is:
Explanation: $\int_{-2}^{2}(4-x^2)dx=32/3$.
Q4.The integrand for area between curves should be:
Explanation: Upper minus lower keeps it non-negative.
Practice more MCQs → 📝 Assignment · 35 marks