What is the integral from 0 to 1 of 16t^3 ?

The integral from 0 to 1 of 16t^3 is 4

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integral from 0 to 1 of 16t^3

\int_{0}^{1} 16 t^{3} d t

4

Solution steps

\int_{0}^{1} 16 t^{3} d t

Take the constant out:

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

= 16 \cdot \int_{0}^{1} t^{3} d t

Apply the Power Rule

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

= 16 [\frac{t ^{4}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= 16 \cdot \frac{1}{4}

Simplify

= 4

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What is the integral from 0 to 1 of 16t^3 ?
The integral from 0 to 1 of 16t^3 is 4