What is the integral from 0 to 1 of (x^3)/2 ?

The integral from 0 to 1 of (x^3)/2 is 1/8

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integral from 0 to 1 of (x^3)/2

\int_{0}^{1} \frac{x ^{3}}{2} d x

\frac{1}{8}

Decimal Notation

0.125

Solution steps

\int_{0}^{1} \frac{x ^{3}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0}^{1}

= \frac{1}{2} [\frac{x ^{4}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= \frac{1}{2} \cdot \frac{1}{4}

Simplify

= \frac{1}{8}

\int_{- 4}^{2} \frac{1}{16 - x ^{2}} d x

\int_{- 1}^{1} - x^{2} d x

\int_{- 3}^{1} (x^{2} + x - 2) d x

\int_{- 2}^{3} \frac{x - 3}{4} d x

\int_{0}^{1} e^{\frac{- x ^{2}}{2}} d x

What is the integral from 0 to 1 of (x^3)/2 ?
The integral from 0 to 1 of (x^3)/2 is 1/8