What is the integral from 0 to 1 of 16x ?

The integral from 0 to 1 of 16x is 8

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integral from 0 to 1 of 16x

\int_{0}^{1} 16 x d x

8

Solution steps

\int_{0}^{1} 16 x d x

Take the constant out:

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

= 16 \cdot \int_{0}^{1} x d x

Apply the Power Rule

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

= 16 [\frac{x ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= 16 \cdot \frac{1}{2}

Simplify

= 8

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