Answer
$\dfrac{4}{15}$
Work Step by Step
The function $\sqrt x (1 - x)$ intersects the graph of $y=0$ (the x axis) when $x=0$ and when $x=1$, therefore we take the integral from $0$ to $1$.
Also, from the graph we can see that the area bounded by the graphs of the equations goes from $x=0$ to $x=1$
$\displaystyle \int_{0}^{1} [\sqrt x (1 - x)]dx = \displaystyle \int_{0}^{1} [x^{\frac{1}{2}} (1 - x)]dx = \displaystyle \int_{0}^{1} [x^{\frac{1}{2}} - x^{\frac{1}{2}}(x)]dx \\
= \displaystyle \int_{0}^{1} [x^{1/2} - x^{\frac{1}{2}+1}]dx = \displaystyle \int_{0}^{1} [x^{\frac{1}{2}} - x^{\frac{3}{2}}]dx = \bigg[ \dfrac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \dfrac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg] \Bigg \rvert_{0}^{1}\\
= \bigg[ \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} - \dfrac{x^{\frac{5}{2}}}{\frac{5}{2}}\bigg] \Bigg \rvert_{0}^{1} = \bigg[ \dfrac{2x^{\frac{3}{2}}}{3} - \dfrac{2x^{\frac{5}{2}}}{5}\bigg] \Bigg \rvert_{0}^{1} \\
= \bigg[ \dfrac{2(1)^{\frac{3}{2}}}{3} - \dfrac{2(1)^{\frac{5}{2}}}{5}\bigg] - \bigg[ \dfrac{2(0)^{\frac{3}{2}}}{3} - \dfrac{2(0)^{\frac{5}{2}}}{5}\bigg] = \bigg[ \dfrac{2(1)}{3} - \dfrac{2(1)}{5}\bigg] - \bigg[ 0 - 0\bigg] \\
= \dfrac{2}{3} - \dfrac{2}{5} - 0 = \dfrac{2}{3} - \dfrac{2}{5} = \dfrac{4}{15}$
