What is the integral from 1/2 to 1 of (x^{-3}-5) ?

The integral from 1/2 to 1 of (x^{-3}-5) is -1

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integral from 1/2 to 1 of (x^{-3}-5)

\int_{\frac{1}{2}}^{1} (x^{- 3} - 5) d x

- 1

Solution steps

\int_{\frac{1}{2}}^{1} x^{- 3} - 5 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} 5 d x

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

\int_{\frac{1}{2}}^{1} 5 d x = \frac{5}{2}

= \frac{3}{2} - \frac{5}{2}

Simplify

\frac{3}{2} - \frac{5}{2} : - 1

= - 1

y^{^{''}} = b y

\int_{1}^{2} \frac{x ^{2} + 5}{3 x - x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2}}{4 + x})

\int_{0}^{\frac{3}{4}} 60 x d x

y^{^{'}} - 3 y + 2 x = 0

What is the integral from 1/2 to 1 of (x^{-3}-5) ?
The integral from 1/2 to 1 of (x^{-3}-5) is -1