What is the integral from 1/2 to 1 of (x^{-3}-10) ?

The integral from 1/2 to 1 of (x^{-3}-10) is -7/2

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integral from 1/2 to 1 of (x^{-3}-10)

\int_{\frac{1}{2}}^{1} (x^{- 3} - 10) d x

- \frac{7}{2}

Decimal

- 3.5

Solution steps

\int_{\frac{1}{2}}^{1} x^{- 3} - 10 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} 10 d x

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

\int_{\frac{1}{2}}^{1} 10 d x = 5

= \frac{3}{2} - 5

Simplify

\frac{3}{2} - 5 : - \frac{7}{2}

= - \frac{7}{2}

\frac{d}{d x} (5 sin (x) + 6 cos (x))

\frac{d}{d x} (1 - \frac{x}{2})

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 6, y (1) = 3

n = 0 \sum \infty \frac{2 ^{n}}{n} x^{n}

\int sin (2 x) sec (x) d x

What is the integral from 1/2 to 1 of (x^{-3}-10) ?
The integral from 1/2 to 1 of (x^{-3}-10) is -7/2