What is the integral from 1 to 3 of 6/(x^3) ?

The integral from 1 to 3 of 6/(x^3) is 8/3

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integral from 1 to 3 of 6/(x^3)

\int_{1}^{3} \frac{6}{x ^{3}} d x

\frac{8}{3}

Solution steps

\int_{1}^{3} \frac{6}{x ^{3}} d x

Take the constant out:

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

= 6 \cdot \int_{1}^{3} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{3}

= 6 [- \frac{1}{2 x ^{2}}]_{1}^{3}

Compute the boundaries:

\frac{4}{9}

= 6 \cdot \frac{4}{9}

Simplify

= \frac{8}{3}

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What is the integral from 1 to 3 of 6/(x^3) ?
The integral from 1 to 3 of 6/(x^3) is 8/3