How to find the area enclosed by a parabola and a line with the y-axis

Expert intro: We first identify which curve is on top, then integrate the vertical gap over the interval cut off by the y-axis. The setup is clean once the intersection point is used.

Detailed walkthrough

Step 1: Identify the boundary on the left

The region starts at the y-axis, so the left endpoint is $x=0$.

The two curves intersect at $(2,4)$, so the right endpoint is $x=2$.

Step 2: Determine top minus bottom

For $0\le x\le 2$:

• upper curve: $y=8-2x$
• lower curve: $y=\frac12 x^2$

Step 3: Integrate the vertical distance

$A=\int_0^2\left[(8-2x)-\frac12 x^2\right]dx$

$A=\int_0^2 \left(8-2x-\frac12 x^2\right)dx$

$A=\left[8x-x^2-\frac{x^3}{6}\right]_0^2$

$A=16-4-\frac{8}{6}=12-\frac43=\frac{32}{3}$

Answer

$\boxed{\frac{32}{3}}$

💡 Pitfall guide

A common mistake is flipping the curves and subtracting the parabola from the line the wrong way around. Another easy slip is using $x=4$ instead of $x=2$; the y-axis, not the x-axis, closes region $R$.

🔄 Real-world variant

If the left boundary were changed from the y-axis to the x-axis, the setup would no longer be the same region. You would have to split the area where the top boundary changes, and the integral would be over a different interval.

🔍 Related terms

definite integral, area between curves, intersection point