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.

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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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.

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.

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.

Log in to reply

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