What is the integral from 8 to infinity of 2x^{-3} ?

The integral from 8 to infinity of 2x^{-3} is 1/64

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integral from 8 to infinity of 2x^{-3}

\int_{8}^{\infty} 2 x^{- 3} d x

\frac{1}{64}

Decimal Notation

0.015625

Solution steps

\int_{8}^{\infty} 2 x^{- 3} d x

Compute the indefinite integral:

\int 2 x^{- 3} d x = - \frac{1}{x ^{2}} + C

Compute the boundaries:

\int_{8}^{\infty} 2 x^{- 3} d x = 0 - (- \frac{1}{64})

= 0 - (- \frac{1}{64})

Simplify

= \frac{1}{64}

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What is the integral from 8 to infinity of 2x^{-3} ?
The integral from 8 to infinity of 2x^{-3} is 1/64