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

The integral from-2 to-4 of (-4)^{-3} is 1/32

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

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

\frac{1}{32}

Decimal Notation

0.03125

Solution steps

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

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

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

Integral of a constant:

\int a d x = a x

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

Simplify

- [(- 4)^{- 3} x]_{- 4}^{- 2} : - [- \frac{1}{64} x]_{- 4}^{- 2}

= - [- \frac{1}{64} x]_{- 4}^{- 2}

Compute the boundaries:

- \frac{1}{32}

= - (- \frac{1}{32})

Simplify

= \frac{1}{32}

\int_{- 1}^{1} x d x

f (t) = t^{3} e^{2 t}

\frac{d}{d x} (- 2 + 6 sin (6 x))

ln (x^{2})

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

What is the integral from-2 to-4 of (-4)^{-3} ?
The integral from-2 to-4 of (-4)^{-3} is 1/32