What is the integral from 6 to 9 of (1/(x^2)) ?

The integral from 6 to 9 of (1/(x^2)) is 1/18

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integral from 6 to 9 of (1/(x^2))

\int_{6}^{9} (\frac{1}{x ^{2}}) d x

\frac{1}{18}

Decimal

0.05555 \dots

Solution steps

\int_{6}^{9} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{6}^{9}

= [- \frac{1}{x}]_{6}^{9}

Compute the boundaries:

\frac{1}{18}

= \frac{1}{18}

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What is the integral from 6 to 9 of (1/(x^2)) ?
The integral from 6 to 9 of (1/(x^2)) is 1/18