How do you evaluate $\int \frac{1}{x^{2}}$ $int 1/x^2$ for [-1, -1/2]?

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How do you evaluate $\int \frac{1}{x^{2}}$ $int 1/x^2$ for [-1, -1/2]?

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Answer 1 · 2015-10-23T02:22:57

We can use the power rule.

$\int \frac{1}{x^{2}} d x = \int x^{- 2} d x$ $int 1/x^2dx = int x^(-2)dx$

$\int_{a}^{b} x^{n} d x = [\frac{x^{n + 1}}{n + 1}] ∣ ∣ ∣$ $int_a^b x^ndx = [(x^(n+1))/(n+1)]|$ $_{a}^{b}$ $""_(a)^(b)$

$= [\frac{b^{n + 1}}{n + 1}] - [\frac{a^{n + 1}}{n + 1}]$ $= [(b^(n+1))/(n+1)] - [(a^(n+1))/(n+1)]$

So, let's use this formula to do so.

$\Rightarrow \int_{- 1}^{-1/2} \frac{1}{x^{2}} d x = \int_{- 1}^{-1/2} x^{- 2} d x$ $=> color(blue)(int_(-1)^("-1/2") 1/x^2dx) = int_(-1)^("-1/2") x^(-2)dx$

$= [\frac{x^{- 1}}{- 1}] ∣_{- 1}^{-1/2} =$ $= [x^(-1)/(-1)]|_(-1)^"-1/2" =$ $[- \frac{1}{x}]$ $[-1/x]$ $∣_{- 1}^{-1/2}$ $|_(-1)^"-1/2"$

$= [- \frac{1}{- 1/2}] - [- \frac{1}{- 1}]$ $= [-1/(-"1/2")] - [-1/(-1)]$

$= 2 - 1$ $= 2 - 1$

$= 1$ $= color(blue)(1)$

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How do you evaluate $\int \frac{1}{x^{2}}$ $int 1/x^2$ for [-1, -1/2]?

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