What is the integral from 6 to infinity of 2x^{-3} ?

The integral from 6 to infinity of 2x^{-3} is 1/36

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integral from 6 to infinity of 2x^{-3}

\int_{6}^{\infty} 2 x^{- 3} d x

\frac{1}{36}

Decimal Notation

0.02777 \dots

Solution steps

\int_{6}^{\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_{6}^{\infty} 2 x^{- 3} d x = 0 - (- \frac{1}{36})

= 0 - (- \frac{1}{36})

Simplify

= \frac{1}{36}

\int_{0}^{5} 0.5 e^{- 0.5 x} d x

\int_{- 2}^{2} \frac{1}{2} (1 - \frac{∣ x ∣}{2}) d x

\int_{0}^{5} \frac{7}{3} x d x

\int_{0}^{6} 0 - (x^{2} - 6 x) d x

\int_{0}^{1} \frac{2}{3 x} d x

What is the integral from 6 to infinity of 2x^{-3} ?
The integral from 6 to infinity of 2x^{-3} is 1/36