What is \(f(x)\)? As currently stated, \(f(x) = 1 + x^2 + x^3 + \ldots\) which doesn't lend itself to a nice pattern to continue the ellipsis.
–
Ivan Koswara
·
4 years ago

Log in to reply

I'm awful in calculus, but I think you can separate the expression on infinite integrals, according to the property that says: the integral of a sum is the sum of the integrals. However, I can't see a clear way to solve it, indeed I think there is a solution.
–
Leonardo Cidrão
·
4 years ago

Log in to reply

Here is what I think:

Rewrite the integrand as

\((1 + x + x^2 + x^3 + ...) - x\)

then we can rewrite it as

\(\frac{1}{1 - x} - x\)

Thus the integrand becomes

\(\int^{3}_2{\frac{1}{1 - x} - x}\)

Can you take it from there? From here it's trivial.
–
Joshua Siktar
·
4 years ago

Log in to reply

@Joshua Siktar
–
This approach is completely invalid. \(\displaystyle \sum_{i=0}^{\infty} x^i = \frac{1}{1-x}\) only when \(|x| < 1\). Thus this definite integral is actually infinity.
–
Bob Krueger
·
4 years ago

## Comments

Sort by:

TopNewestWhat is \(f(x)\)? As currently stated, \(f(x) = 1 + x^2 + x^3 + \ldots\) which doesn't lend itself to a nice pattern to continue the ellipsis. – Ivan Koswara · 4 years ago

Log in to reply

I'm awful in calculus, but I think you can separate the expression on infinite integrals, according to the property that says: the integral of a sum is the sum of the integrals. However, I can't see a clear way to solve it, indeed I think there is a solution. – Leonardo Cidrão · 4 years ago

Log in to reply

Here is what I think:

Rewrite the integrand as

\((1 + x + x^2 + x^3 + ...) - x\)

then we can rewrite it as

\(\frac{1}{1 - x} - x\)

Thus the integrand becomes

\(\int^{3}_2{\frac{1}{1 - x} - x}\)

Can you take it from there? From here it's trivial. – Joshua Siktar · 4 years ago

Log in to reply

– Bob Krueger · 4 years ago

This approach is completely invalid. \(\displaystyle \sum_{i=0}^{\infty} x^i = \frac{1}{1-x}\) only when \(|x| < 1\). Thus this definite integral is actually infinity.Log in to reply