Note that

and use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove that for positive :

Using compute the integral to six decimal places.

Recall the weighted mean value theorem:

For functions and continuous on , if never changes sign in then there exists such that

* Proof. * Let

Then,

for some . Since is strictly decreasing on , we know ; thus,

Furthermore,

Thus,

Now, taking we compute,

there is an error it must be f(c) not 1/c