What is the integral from 0 to 3 of 4t ?

The integral from 0 to 3 of 4t is 18

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integral from 0 to 3 of 4t

\int_{0}^{3} 4 t d t

18

Solution steps

\int_{0}^{3} 4 t d t

Take the constant out:

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

= 4 \cdot \int_{0}^{3} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{3}

= 4 [\frac{t ^{2}}{2}]_{0}^{3}

Compute the boundaries:

\frac{9}{2}

= 4 \cdot \frac{9}{2}

Simplify

= 18

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