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.
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Ivan Koswara
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3 years, 6 months ago

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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.
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Leonardo Cidrão
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3 years, 6 months ago

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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.
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Joshua Siktar
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3 years, 6 months ago

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@Joshua Siktar
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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.
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Bob Krueger
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3 years, 6 months ago

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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 · 3 years, 6 months ago

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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 · 3 years, 6 months ago

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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 · 3 years, 6 months ago

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– Bob Krueger · 3 years, 6 months 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