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

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

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

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

0

Solution steps

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

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

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

= 3 - 3

Simplify

= 0

\int_{0}^{6} x d x

\int_{- 1}^{1} (2 x^{2} + 1) d x

\int_{1}^{2} \frac{x ^{2} + 3}{x} d x

\int_{0}^{1} 1 + x^{2} d x

\int_{0}^{2} x^{2} - 3 x + 2 d x

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