What is the integral from 1/3 to 3 of (4-4/(x^2)) ?

The integral from 1/3 to 3 of (4-4/(x^2)) is 0

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integral from 1/3 to 3 of (4-4/(x^2))

\int_{\frac{1}{3}}^{3} (4 - \frac{4}{x ^{2}}) d x

0

Solution steps

\int_{\frac{1}{3}}^{3} 4 - \frac{4}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{\frac{1}{3}}^{3} 4 d x - \int_{\frac{1}{3}}^{3} \frac{4}{x ^{2}} d x

\int_{\frac{1}{3}}^{3} 4 d x = \frac{32}{3}

\int_{\frac{1}{3}}^{3} \frac{4}{x ^{2}} d x = \frac{32}{3}

= \frac{32}{3} - \frac{32}{3}

Simplify

= 0

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What is the integral from 1/3 to 3 of (4-4/(x^2)) ?
The integral from 1/3 to 3 of (4-4/(x^2)) is 0