What is the integral from 0 to 1 of 16 1/(x^p) ?

The integral from 0 to 1 of 16 1/(x^p) is (16)/(-p+1)

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integral from 0 to 1 of 16 1/(x^p)

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

\frac{16}{- p + 1}

Solution steps

\int_{0}^{1} 16 \cdot \frac{1}{x ^{p}} 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} \frac{1}{x ^{p}} d x

Apply the Power Rule

= 16 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{- p + 1}

= 16 \cdot \frac{1}{- p + 1}

Simplify

= \frac{16}{- p + 1}

What is the integral from 0 to 1 of 16 1/(x^p) ?
The integral from 0 to 1 of 16 1/(x^p) is (16)/(-p+1)