Question

The area (in sq. units) of the region described by {(x, y) : y^2 < 2x, and y > 4x –1} is

Answer 1

9/32

Answer 2

The region is bounded by the parabola $y^2=2x$ and the line $y=4x-1$. We rewrite the equations as $x = y^2/2$ and $x = (y+1)/4$. The intersection points are found by setting these equal, yielding $y=1$ and $y=-1/2$. Integrating with respect to $y$ from $-1/2$ to $1$, the area is the integral of the difference between the right curve ($x_{line}$) and the left curve ($x_{parabola}$). The integral $\int_{-1/2}^{1} \left(\frac{y+1}{4} - \frac{y^2}{2}\right) dy$ evaluates to $\frac{9}{32}$.