Calculus II Average Value of a Function

Question

The average value of the function v(x)=8/x² on the interval [1,c] is equal to 1.

Find the value of c.

Answer 1

Average value of f(x) on the interval [a,b] is 1/(b-a)∫_{(from a to b)}f(x)dx

So, we have 1 / (c - 1)∫_{(from 1 to c)}(8x^-2)dx = 1

1/(c-1)[-8/x]_{(from 1 to c)} = 1

1/(c-1)[-8/c + 8] = 1

1/(c-1)[(-8 + 8c)/c] = 1

-8 + 8c = c(c-1)

c² - 9c + 8 = 0

(c - 8)(c - 1) = 0

c = 8 or c = 1

Since c = 1 is not possible, c = 8.