What is the integral from-4 to-6 of (-2)^{-3} ?

The integral from-4 to-6 of (-2)^{-3} is 1/4

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integral from-4 to-6 of (-2)^{-3}

\int_{- 4}^{- 6} (- 2)^{- 3} d x

\frac{1}{4}

Decimal

0.25

Solution steps

\int_{- 4}^{- 6} (- 2)^{- 3} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 6}^{- 4} (- 2)^{- 3} d x

Integral of a constant:

\int a d x = a x

= - [(- 2)^{- 3} x]_{- 6}^{- 4}

Simplify

- [(- 2)^{- 3} x]_{- 6}^{- 4} : - [- \frac{1}{8} x]_{- 6}^{- 4}

= - [- \frac{1}{8} x]_{- 6}^{- 4}

Compute the boundaries:

- \frac{1}{4}

= - (- \frac{1}{4})

Simplify

= \frac{1}{4}

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What is the integral from-4 to-6 of (-2)^{-3} ?
The integral from-4 to-6 of (-2)^{-3} is 1/4