The area bounded by the parabola y^2=4x and the line 2x-3y+4=0, in square units, is:

Question

The area bounded by the parabola $y^2=4x$ and the line $2x-3y+4=0$, in square units, is:

Accepted Answer

The curves meet at $y=1,4$ (solving $2\left(\dfrac{y^2}{4}\right)-3y+4=0 \Rightarrow y^2-6y+8=0$).

$$\text{Area} = \int_1^4\left(\frac{3y-4}{2}-\frac{y^2}{4}\right)dy = \left[\frac{3y^2}{4}-2y-\frac{y^3}{12}\right]_1^4 = \frac13$$

Solutions · 1