What is the integral from 0 to 1 of 3y^3 ?

The integral from 0 to 1 of 3y^3 is 3/4

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integral from 0 to 1 of 3y^3

\int_{0}^{1} 3 y^{3} d y

\frac{3}{4}

Decimal

0.75

Solution steps

\int_{0}^{1} 3 y^{3} d y

Take the constant out:

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

= 3 \cdot \int_{0}^{1} y^{3} d y

Apply the Power Rule

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

= 3 [\frac{y ^{4}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= 3 \cdot \frac{1}{4}

Simplify

= \frac{3}{4}

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What is the integral from 0 to 1 of 3y^3 ?
The integral from 0 to 1 of 3y^3 is 3/4