What is the integral from 1/2 to 1 of (x^{-3}-19) ?

The integral from 1/2 to 1 of (x^{-3}-19) is -8

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integral from 1/2 to 1 of (x^{-3}-19)

\int_{\frac{1}{2}}^{1} (x^{- 3} - 19) d x

- 8

Solution steps

\int_{\frac{1}{2}}^{1} x^{- 3} - 19 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}}^{1} x^{- 3} d x - \int_{\frac{1}{2}}^{1} 19 d x

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

\int_{\frac{1}{2}}^{1} 19 d x = \frac{19}{2}

= \frac{3}{2} - \frac{19}{2}

Simplify

\frac{3}{2} - \frac{19}{2} : - 8

= - 8

What is the integral from 1/2 to 1 of (x^{-3}-19) ?
The integral from 1/2 to 1 of (x^{-3}-19) is -8