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

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

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

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

0

Solution steps

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

\int_{\frac{1}{2}}^{2} 4 d x = 6

\int_{\frac{1}{2}}^{2} \frac{4}{x ^{2}} d x = 6

= 6 - 6

Simplify

= 0

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

\int_{2}^{4} (2 x + 1) d x

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 4)^{2} d x

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

\int_{2}^{6} \frac{1}{x ^{2}} d x

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