How to find the area enclosed by two curves and the x-axis

Key concept: Region S changes its top boundary at the intersection point, so the area has to be split into two pieces. That is the whole trick here.

Step by step

Step 1: Split the region

Region $S$ is bounded by the curves and the x-axis.

• On $0\le x\le 2$, the top curve is $y=\frac12 x^2$
• On $2\le x\le 4$, the top curve is $y=8-2x$

Step 2: Write the area as two integrals

$A=\int_0^2 \frac12 x^2\,dx+\int_2^4 (8-2x)\,dx$

Step 3: Compute each part

$\int_0^2 \frac12 x^2\,dx=\left[\frac{x^3}{6}\right]_0^2=\frac{8}{6}=\frac43$

$\int_2^4 (8-2x)\,dx=\left[8x-x^2\right]_2^4=(32-16)-(16-4)=4$

Step 4: Add them

$A=\frac43+4=\frac{16}{3}$

Answer

$\boxed{\frac{16}{3}}$

Pitfall alert

The main trap is trying to use one single formula for the whole interval. The upper boundary changes at $x=2$, so the integral must be split there.

Try different conditions

If the x-axis were not part of the boundary, you would instead find the area between the line and parabola only, which gives a different integral over $[0,2]$. The lower boundary is what changes the setup most.

Further reading

piecewise integral, bounded region, area under a curve