What is the integral from 2 to 5 of (10)/(9x^2) ?

The integral from 2 to 5 of (10)/(9x^2) is 1/3

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integral from 2 to 5 of (10)/(9x^2)

\int_{2}^{5} \frac{10}{9 x ^{2}} d x

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

\int_{2}^{5} \frac{10}{9 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{10}{9} \cdot \int_{2}^{5} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{2}^{5}

= \frac{10}{9} [- \frac{1}{x}]_{2}^{5}

Compute the boundaries:

\frac{3}{10}

= \frac{10}{9} \cdot \frac{3}{10}

Simplify

= \frac{1}{3}

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What is the integral from 2 to 5 of (10)/(9x^2) ?
The integral from 2 to 5 of (10)/(9x^2) is 1/3