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

The integral from 2 to infinity of 8x^{-2} is 4

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

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

4

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{2}^{\infty} 8 x^{- 2} d x = 0 - (- 4)

= 0 - (- 4)

Simplify

= 4

\int 7 - \frac{1}{7 x ^{2}} d x

x \to 0 lim ((1 - e^{x})^{x})

\int \frac{x ^{6}}{( 8 - x ^{7} ) ^{2}} d x

\frac{1}{x - 1}, 6

x \to 0 lim (\frac{e ^{x} - 1}{18 x})

What is the integral from 2 to infinity of 8x^{-2} ?
The integral from 2 to infinity of 8x^{-2} is 4