What is the integral from 1 to infinity of x^{-9} ?

The integral from 1 to infinity of x^{-9} is 1/8

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integral from 1 to infinity of x^{-9}

\int_{1}^{\infty} x^{- 9} d x

\frac{1}{8}

Decimal

0.125

Solution steps

\int_{1}^{\infty} x^{- 9} d x

Compute the indefinite integral:

\int x^{- 9} d x = - \frac{1}{8 x ^{8}} + C

Compute the boundaries:

\int_{1}^{\infty} x^{- 9} d x = 0 - (- \frac{1}{8})

= 0 - (- \frac{1}{8})

Simplify

= \frac{1}{8}

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

f (x) = (x + 1) (- \frac{1}{x + 2})

\frac{\partial}{\partial x} (3 arctan (x y^{2}))

(cos (θ))^{^{'}}

\int_{- 1}^{3} (u^{3} + u) d u

What is the integral from 1 to infinity of x^{-9} ?
The integral from 1 to infinity of x^{-9} is 1/8