How do you evaluate the definite integral $\int (\frac{2}{r^{3}}) d r$ $int (2/r^3)dr$ from $[- 3, - 1]$ $[-3,-1]$ ?

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How do you evaluate the definite integral $\int (\frac{2}{r^{3}}) d r$ $int (2/r^3)dr$ from $[- 3, - 1]$ $[-3,-1]$ ?

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Answer 1 · 2017-08-10T06:03:01

The integral equals $- \frac{8}{9}$ $-8/9$

Explanation:

Use the basic power rule.

$\int_{- 3}^{- 1} \frac{2}{r^{3}} d r = {[- r^{- 2}]}_{- 3}^{- 2} = - {(- 1)}^{- 2} - (- {(- 3)}^{- 2}) = - 1 + \frac{1}{9} = - \frac{8}{9}$ $int_(-3)^(-1) 2/r^3 dr = [-r^-2]_(-3)^-1 = -(-1)^-2 - (-(-3)^-2) =-1 + 1/9 = -8/9$

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How do you evaluate the definite integral $\int (\frac{2}{r^{3}}) d r$ $int (2/r^3)dr$ from $[- 3, - 1]$ $[-3,-1]$ ?

1 Answer

Explanation:

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How do you evaluate the definite integral $\int (\frac{2}{r^{3}}) d r$ $int (2/r^3)dr$ from $[- 3, - 1]$ $[-3,-1]$ ?