Question

Suppose f is continuous, f(0) = 0, f(1) = 1, f'(x) > 0 and $\int_{0}^{1} f(x)dx = \frac{1}{3}$. Find the value of the definite integral $\int_{0}^{1} f^{-1}(y)dy$.

Answer 1

$\frac{2}{3}$

Answer 2

Let $I = \int_{0}^{1} f^{-1}(y)dy$. We use the substitution $y = f(x)$. Since $f(0) = 0$ and $f(1) = 1$, the limits of integration for $x$ are from 0 to 1. Differentiating $y = f(x)$ with respect to $x$, we get $dy = f'(x)dx$. From $y = f(x)$, we have $x = f^{-1}(y)$. Substituting these into the integral $I$: $I = \int_{0}^{1} x \cdot f'(x) dx$ Now, we apply integration by parts. Let $u = x$ and $dv = f'(x)dx$. Then, $du = dx$ and $v = f(x)$. Applying the integration by parts formula $\int u \, dv = uv - \int v \, du$: $I = [x f(x)]_{0}^{1} - \int_{0}^{1} f(x) dx$ Evaluate the term $[x f(x)]_{0}^{1}$: $[x f(x)]_{0}^{1} = (1 \cdot f(1)) - (0 \cdot f(0))$ Given $f(1) = 1$ and $f(0) = 0$: $[x f(x)]_{0}^{1} = (1 \cdot 1) - (0 \cdot 0) = 1$ Substitute this value back into the expression for $I$: $I = 1 - \int_{0}^{1} f(x) dx$ Given that $\int_{0}^{1} f(x)dx = \frac{1}{3}$: $I = 1 - \frac{1}{3}$ $I = \frac{2}{3}$