Question

For an odd function f(x) where integral^-5_0 f(x) dx=3 determine the average value of f on the interval [-5,5]. 0 3/10 -3/10 15/2 -3/5

Answer 1

Since $f(x)$ is an odd function, we have $f(x) = -f(-x)$ for all $x$ in the domain of $f(x)$. Therefore,

\begin{align*}

\int_{-5}^5 f(x) dx &= \int_{-5}^0 f(x) dx + \int_0^5 f(x) dx \

&= \int_{5}^0 -f(-x) dx + \int_0^5 f(x) dx &\quad\text{(using substitution)} \

&= \int_{0}^5 f(-x) dx + \int_0^5 f(x) dx \

&= 2\int_0^5 f(x) dx \

&= 2\cdot \frac{1}{5}\int_{-5}^5 f(x) dx \

&= 2\cdot \frac{1}{5} \cdot 3 \

&= \frac{6}{5}.

\end{align*}

Thus, the average value of $f$ on the interval $[-5,5]$ is $\frac{1}{10} \int_{-5}^5 f(x) dx = \frac{6}{5}\cdot\frac{1}{10} = \boxed{\frac{3}{5}}$.