Answer
$\int_0^1f^{-1}(y)dy=\frac{2}{3}$
Work Step by Step
Let $x=f^{-1}(y)$ or $y=f(x)$.
Then,
i) $\frac{dy}{dx}=f'(x)$ or $dy=f'(x)dx$;
ii) for $y_1=0\Rightarrow x_1=f^{-1}(y_1)=f^{-1}(0)=0$ (since $f(0)=0\Leftrightarrow f^{-1}(1)=1$);
iii) for $y_2=1\Rightarrow x_2=f^{-1}(y_2)=f^{-1}(1)=1$ (since $f(1)=1\Leftrightarrow f^{-1}(1)=1$).
Using the substitution rule,
$\int_{y_1=0}^{y_2=1}f^{-1}(y)dy=\int_{x_1=0}^{x_2=1}xf'(x)dx$
$=\int_{0}^{1}(xf'(x)+f(x))-f(x)dx$ (Apply the Difference property for integrals)
$=\int_0^1(xf'(x)+f(x))dx-\int_0^1f(x)dx$
$=[xf(x)]_0^1-\frac{1}{3}$ (Since $\frac{d}{dx}(xf(x))=xf'(x)+f(x)$)
$=1\cdot f(1)-0\cdot f(0)-\frac{1}{3}$
$=1\cdot 1-0\cdot 0-\frac{1}{3}$
$=\frac{2}{3}$
Thus,
$\int_0^1f^{-1}(y)dy=\frac{2}{3}$