How Do You Find the Area Between Two Curves?

Find the intersection points, identify which curve is above the other, then integrate upper curve minus lower curve between the intersections. If the curves change order, split the area into separate integrals.

The Method

For curves $y=f(x)$ and $y=g(x)$, the area from $x=a$ to $x=b$ is

$$A=\int_{a}^{b}\left(f(x)-g(x)\right)\,dx$$

provided $f(x)\geq g(x)$ throughout that interval. A definite integral measures signed area, so the subtraction order matters. When $f(x)$ is above $g(x)$, every value of (f(x)-g(x)) is non-negative, and integration accumulates the vertical separation between the graphs. Equivalently, area can be written as $$, but splitting the integral is often clearer and easier to evaluate analytically.

A reliable process is:

1. Solve $f(x)=g(x)$ to find the points of intersection and hence the integration limits.
2. Determine the upper curve, using the graph or a test value between the intersections.
3. Write the correct integral before evaluating it.
4. If the curves exchange positions, split the interval where they cross.

Worked Example

Find the area enclosed by $y=2x$ and $y=x^2$.

First find the intersections:

$$x^2=2x\Rightarrow x(x-2)=0\Rightarrow x=0,2.$$

The intersections provide the limits $0$ and $2$. For example, at $x=1$, $2x=2$ while $x^2=1$, confirming that the line is the upper curve. This test prevents a reversed subtraction. Therefore,

$$A=\int_0^2(2x-x^2)\,dx =\left[x^2-\frac{x^3}{3}\right]_0^2 =4-\frac{8}{3} =\frac{4}{3}.$$

Thus, the enclosed area is $\frac{4}{3}$ square units.

IB Exam Technique

This is IB Mathematics AA SL 5.11 and can be tested analytically on Paper 1 or with GDC-supported evaluation on Paper 2. On Paper 2, a GDC may find intersections and evaluate the integral, but you must still record the equations solved, suitable limits, and the integral itself. Give an exact answer when available; otherwise use three significant figures unless instructed differently.

Examiners expect the correct integral expression before its value. A common misconception is to integrate one curve or subtract in the wrong order: a definite integral gives signed area, but geometric area must be non-negative.