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

The integral from 1 to 2 of 3/(2x^3) is 9/16

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

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

\frac{9}{16}

Decimal Notation

0.5625

Solution steps

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

Take the constant out:

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

= \frac{3}{2} \cdot \int_{1}^{2} \frac{1}{x ^{3}} d x

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{8}

= \frac{3}{2} \cdot \frac{3}{8}

Simplify

= \frac{9}{16}

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What is the integral from 1 to 2 of 3/(2x^3) ?
The integral from 1 to 2 of 3/(2x^3) is 9/16