What is the integral from 4 to infinity of 8x^{-3} ?

The integral from 4 to infinity of 8x^{-3} is 1/4

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integral from 4 to infinity of 8x^{-3}

\int_{4}^{\infty} 8 x^{- 3} d x

\frac{1}{4}

Decimal

0.25

Solution steps

\int_{4}^{\infty} 8 x^{- 3} d x

Compute the indefinite integral:

\int 8 x^{- 3} d x = - \frac{4}{x ^{2}} + C

Compute the boundaries:

\int_{4}^{\infty} 8 x^{- 3} d x = 0 - (- \frac{1}{4})

= 0 - (- \frac{1}{4})

Simplify

= \frac{1}{4}

What is the integral from 4 to infinity of 8x^{-3} ?
The integral from 4 to infinity of 8x^{-3} is 1/4