What is the integral from 0 to t of 8t ?

The integral from 0 to t of 8t is 4t^2

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integral from 0 to t of 8t

\int_{0}^{t} 8 t d t

4 t^{2}

Solution steps

\int_{0}^{t} 8 t d t

Take the constant out:

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

= 8 \cdot \int_{0}^{t} t d t

Apply the Power Rule

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

= 8 [\frac{t ^{2}}{2}]_{0}^{t}

Compute the boundaries:

\frac{t ^{2}}{2}

= 8 \cdot \frac{t ^{2}}{2}

Simplify

= 4 t^{2}

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