5. Areas Under & Between Curves

Video coming soon3 worked examples

Theory

The area under $y = f(x)$ (above the $x$ -axis) from $a$ to $b$ is $\int_a^b y\,dx$ . The area between two curves, with $f$ the upper and $g$ the lower, is:

\text{Area} = \int_a^b \bigl(f(x) - g(x)\bigr)\,dx

where $a$ and $b$ are the points of intersection.

Advanced Higher also asks for the area between a curve and the $y$ -axis. The whole picture simply turns on its side: strips are now horizontal, so we integrate $x$ with respect to $y$ , between $y$ -limits:

\text{Area} = \int_c^d x\,dy

To use it, rearrange the equation of the curve to give $x$ in terms of $y$ before integrating.

The Golden Rule: for the area between curves, find the intersection points first (they are the limits), then integrate “upper minus lower”. If the region is bounded by the $y$ -axis rather than the $x$ -axis, switch to $\int x\,dy$ and use $y$ -limits throughout.

⚠️ Common Examiner Traps

Wrong way round: it is upper curve minus lower curve — check which is on top over the interval.
Limits: the intersection points give the limits of integration — solve $f(x) = g(x)$ first.
Areas below the axis: a region below the $x$ -axis gives a negative integral, so account for the sign.

Worked examples

Example 1

Find the area enclosed between the curve $y = x^2$ and the line $y = 2x$ .

Step 1: Find the intersection points by solving $x^2 = 2x$ :

x^2 - 2x = 0 \implies x(x-2) = 0 \implies x = 0,\ 2

Step 2: Between these, the line $y = 2x$ is above $y = x^2$ , so integrate upper minus lower:

\text{Area} = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2

Step 3: Evaluate:

= 4 - \frac{8}{3} = \frac{4}{3}

Example 2

Find the area under the curve $y = 3x^2$ between $x = 1$ and $x = 2$ .

Step 1: Integrate the function between the given limits:

\text{Area} = \int_1^2 3x^2\,dx = \left[x^3\right]_1^2

Step 2: Evaluate:

= 8 - 1 = 7

Example 3

Find the area enclosed by the curve $y = x^2$ (for $x \ge 0$ ), the $y$ -axis, and the lines $y = 1$ and $y = 4$ .

Step 1: The region is bounded by the $y$ -axis and given $y$ -limits, so integrate with respect to $y$ . Rearrange the curve to give $x$ in terms of $y$ , taking the positive root since $x \ge 0$ :

y = x^2 \implies x = \sqrt{y} = y^{1/2}

Step 2: Set up the integral with the $y$ -limits:

\text{Area} = \int_1^4 y^{1/2}\,dy = \left[\frac{2}{3}y^{3/2}\right]_1^4

Step 3: Evaluate, noting $4^{3/2} = 8$ and $1^{3/2} = 1$ :

= \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}

The area is $\dfrac{14}{3}$ square units.