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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

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

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

= \frac{3}{2} - 1

Simplify

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

= \frac{1}{2}

y = 6 x e^{x}, (0, 0)

cos (4 t)

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

y = x^{2}, y = 2

3 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

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