Question

The average value of a function f over the interval [−1,2] is −4, and the average value of f over the interval [2,7] is 8. What is the average value of f over the interval [−1,7] ?

Answer 1

The average value of a continuous function f(x) over an interval [a, b] is

$\displaystyle f_{\mathrm{ave}[a,b]} = \frac1{b-a}\int_a^b f(x)\,dx$

We're given that

$\displaystyle f_{\rm ave[-1,2]} = \frac13 \int_{-1}^2 f(x) \, dx = -4$

$\displaystyle f_{\rm ave[2,7]} = \frac15 \int_2^7 f(x) \, dx = 8$

and we want to determine

$\displaystyle f_{\rm ave[-1,7]} = \frac18 \int_{-1}^7 f(x) \, dx$

By the additive property of definite integration, we have

$\displaystyle \int_{-1}^7 f(x) \, dx = \int_{-1}^2 f(x)\,dx + \int_2^7 f(x)\,dx$

so it follows that

$\displaystyle f_{\rm ave[-1,7]} = \frac18 \left(\int_{-1}^2 f(x)\,dx + \int_2^7 f(x)\,dx\right)$

$\displaystyle f_{\rm ave[-1,7]} = \frac18 \left(3\times(-4) + 5\times8\right)$

$\displaystyle f_{\rm ave[-1,7]} = \boxed{\frac72}$