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

The integral from 1/2 to 1 of (x^{-3}-3) is 0

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

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

0

Solution steps

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

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

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

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

Simplify

= 0

x \to 2 lim (x^{3} + 2 x^{2} + 1)

\frac{d y}{d x}, x^{3} - 3 x y^{2} + y^{3} = 1

y^{^{'}} = 3 x + y, y (0) = 5

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 13} d x

y = 2 x, (1, 2)

What is the integral from 1/2 to 1 of (x^{-3}-3) ?
The integral from 1/2 to 1 of (x^{-3}-3) is 0