What is the integral from 1/2 to 1 of (x^{-3}-6) ?

The integral from 1/2 to 1 of (x^{-3}-6) is -3/2

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integral from 1/2 to 1 of (x^{-3}-6)

\int_{\frac{1}{2}}^{1} (x^{- 3} - 6) d x

- \frac{3}{2}

Decimal Notation

- 1.5

Solution steps

\int_{\frac{1}{2}}^{1} x^{- 3} - 6 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} 6 d x

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

\int_{\frac{1}{2}}^{1} 6 d x = 3

= \frac{3}{2} - 3

Simplify

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

= - \frac{3}{2}

3 x d x + 4 y d y = 0

\frac{d}{d x} (\frac{x}{x + 9})

\frac{d}{d x} ((x^{2} - 9)^{2})

(- 5, 5), (5, - 2)

What is the integral from 1/2 to 1 of (x^{-3}-6) ?
The integral from 1/2 to 1 of (x^{-3}-6) is -3/2