What is the integral from 3 to 9 of (1/(x^2)) ?

The integral from 3 to 9 of (1/(x^2)) is 2/9

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integral from 3 to 9 of (1/(x^2))

\int_{3}^{9} (\frac{1}{x ^{2}}) d x

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

\int_{3}^{9} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{3}^{9}

= [- \frac{1}{x}]_{3}^{9}

Compute the boundaries:

\frac{2}{9}

= \frac{2}{9}

\int_{- 2}^{4} (x + 2) d x

\int_{0}^{3} \frac{1}{27} (x + y) d y

\int_{4}^{16} x - 1 d x

\int_{x}^{2 x} \frac{x}{y} d y

\int_{0.4}^{0.6} \frac{5}{3} d x

What is the integral from 3 to 9 of (1/(x^2)) ?
The integral from 3 to 9 of (1/(x^2)) is 2/9