Question

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

Short answer

Answer: The value of the integral is \(\frac{2}{3}\).

Step-by-step solution

Apply Inverse Function Properties

Since f is both continuous and increasing (due to \(f^{\prime}(x) > 0\)), it has an inverse function. We can express the function \(\mathrm{f}^{-1}(\mathrm{y})\) as the inverse of the function f, which means \(f(f^{-1}(y))=y\). Also, note that \(f^{-1}(0)=0\) and \(f^{-1}(1)=1\), since \(f(0)=0\) and \(f(1)=1\). So the integral becomes: $$ \int_{0}^{1} \mathrm{f}^{-1}(\mathrm{y}) \mathrm{dy} $$

Change of Variables

Now, let's apply a change of variables. Introduce a new variable \(x=f^{-1}(y)\). Then \(y = f(x)\) and also \(dy = f'(x)dx\). Now the integral becomes: $$ \int_{0}^{1} x f'(x) dx $$ Notice that the limits didn't change since \(f^{-1}(0)=0\) and \(f^{-1}(1)=1\).

Integration by Parts

To solve the new integral, we can use integration by parts. We let \(u=x\) and \(dv = f'(x)dx\), so that \(du=dx\) and \(v=f(x)\). The formula for integration by parts states: $$ \int u dv = uv - \int v du $$ Applying this formula to our integral, we get: $$ \int_{0}^{1} x f'(x) dx = \left[xf(x)\right]_{0}^{1} - \int_{0}^{1} f(x) dx $$

Evaluate the Integral

Now, we can evaluate the terms in the expression: $$ \left[xf(x)\right]_{0}^{1} = (1\cdot f(1)) - (0\cdot f(0)) = 1 - 0 = 1 $$ The right-hand side integral is given as: $$ \int_{0}^{1} f(x) dx = \frac{1}{3} $$

Calculate the Final Value

Now we can plug these values back into the expression from Step 3: $$ \int_{0}^{1} x f'(x) dx = 1 - \frac{1}{3} $$ So, the value of the integral is: $$ \int_{0}^{1} \mathrm{f}^{-1}(\mathrm{y}) \mathrm{dy} = \frac{2}{3} $$