Average Function Value — Question 4

Question 4

Find the average value of the function $$f(x)=\sqrt{x}(1+x)$$ on the interval $[0,1]$.

Question 4 - Solution

The average value of a function on $[a,b]$ is $$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.$$

Here, $$f_{\text{avg}}=\int_{0}^{1} \sqrt{x}(1+x)\,dx.$$

First expand the integrand: $$\sqrt{x}(1+x)=x^{1/2}+x^{3/2}.$$

Integrate term by term: $$\int x^{1/2}\,dx=\frac{2}{3}x^{3/2}, \qquad \int x^{3/2}\,dx=\frac{2}{5}x^{5/2}.$$

Thus, $$\int_{0}^{1} \sqrt{x}(1+x)\,dx = \left[\frac{2}{3}x^{3/2}+\frac{2}{5}x^{5/2}\right]_{0}^{1}.$$

Evaluate: $$\frac{2}{3}+\frac{2}{5} = \frac{10+6}{15} = \frac{16}{15}.$$

$$\boxed{\frac{16}{15}}$$