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

The integral from 1/2 to 1 of (x^{-3}-8) is -5/2

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

\int_{\frac{1}{2}}^{1} (x^{- 3} - 8) d x

- \frac{5}{2}

Decimal Notation

- 2.5

Solution steps

\int_{\frac{1}{2}}^{1} x^{- 3} - 8 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} 8 d x

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

\int_{\frac{1}{2}}^{1} 8 d x = 4

= \frac{3}{2} - 4

Simplify

\frac{3}{2} - 4 : - \frac{5}{2}

= - \frac{5}{2}

\frac{\partial}{\partial x} (- 7 e^{y} sin (2 x) + y + 5 x y)

\frac{\partial}{\partial x} (x^{- 7} y^{2} + x y^{2} + 2 x y)

n = 0 \sum \infty (2 n)!

x \to - 2 lim (9 x^{2})

x \to - 3 lim (\frac{x ^{2}}{9 - x ^{2}})

What is the integral from 1/2 to 1 of (x^{-3}-8) ?
The integral from 1/2 to 1 of (x^{-3}-8) is -5/2