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

The integral from 1 to 2 of 1/(u^3) is 3/8

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

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

\frac{3}{8}

Decimal

0.375

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{8}

= \frac{3}{8}

\int_{0}^{1} x \cdot sin (x^{2}) d x

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

What is the integral from 1 to 2 of 1/(u^3) ?
The integral from 1 to 2 of 1/(u^3) is 3/8