What is the integral from-3 to-5 of-4^{-3} ?

The integral from-3 to-5 of-4^{-3} is 1/32

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integral from-3 to-5 of-4^{-3}

\int_{- 3}^{- 5} - 4^{- 3} d x

\frac{1}{32}

Decimal

0.03125

Solution steps

\int_{- 3}^{- 5} - 4^{- 3} d x

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

= - \int_{- 5}^{- 3} - 4^{- 3} d x

Integral of a constant:

\int a d x = a x

= - [(- 4^{- 3}) x]_{- 5}^{- 3}

Simplify

- [(- 4^{- 3}) x]_{- 5}^{- 3} : - [- \frac{1}{64} x]_{- 5}^{- 3}

= - [- \frac{1}{64} x]_{- 5}^{- 3}

Compute the boundaries:

- \frac{1}{32}

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

Simplify

= \frac{1}{32}

\frac{d}{d x} (e^{- 16 x})

n = 1 \sum \infty \frac{6}{π ^{n}}

\int \frac{4}{( x - 2 ) ( x + 2 )} d x

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

\frac{1}{9 s + 1}

What is the integral from-3 to-5 of-4^{-3} ?
The integral from-3 to-5 of-4^{-3} is 1/32