Area — Question 4

Question 4 - Solution

The intersections satisfy $x\sqrt{2-x}=x$, giving $x=0,1$.

For $0<x<1$, the square-root curve is above the line, hence

$$A=\int_0^1\bigl(x\sqrt{2-x}-x\bigr)\,dx.$$

With $u=2-x$,

$$\int_0^1x\sqrt{2-x}\,dx=\int_1^2(2-u)\sqrt u\,du =\left[\frac43u^{3/2}-\frac25u^{5/2}\right]_1^2 =\frac{16\sqrt2-14}{15}.$$

$$\boxed{A=\frac{32\sqrt2-43}{30}>0.}$$