How do you evaluate the integral $int 1/x^3dx$ from -1 to 2?

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Answer 1 · 2016-08-23T12:29:01

$3/8$

$int_-1^2 1/x^3 dx = int_-1^2 x^-3 dx$

$= x^-2/-2|_-1 ^2$

$= -1/(2x^2)|_-1 ^2$

$=-1/8- (-1/2)$

$=3/8$

How do you evaluate the integral int 1/x^3dxint 1/x^3dx from -1 to 2?