Question

Find the exact positive value of c if the average values of the function
v(x) = 9/x^2
on the interval [1, c] is equal to 1.

Answer 1

M = average value of the function from 'a' to 'b'
\[\Large M = \frac{1}{b-a}\int_{a}^{b}f(x)dx\]
\[\Large 1 = \frac{1}{c-1}\int_{1}^{c}\frac{9}{x^2}dx\]
Integrate the right side with respect to x. Then evaluate at the limits of integration and solve for c.