What is the integral from 0 to 1 of 19 1/(x^p) ?

The integral from 0 to 1 of 19 1/(x^p) is (19)/(-p+1)

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integral from 0 to 1 of 19 1/(x^p)

\int_{0}^{1} 19 \frac{1}{x ^{p}} d x

\frac{19}{- p + 1}

Solution steps

\int_{0}^{1} 19 \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

= 19 \cdot \int_{0}^{1} \frac{1}{x ^{p}} d x

Apply the Power Rule

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

= 19 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{- p + 1}

= 19 \cdot \frac{1}{- p + 1}

Simplify

= \frac{19}{- p + 1}

f (x) = 7 x^{6} e^{x} + e^{x} x^{7}

\frac{3}{s ( 6 s ^{2} + 5 s + 1 )}

\frac{\partial}{\partial y} (cos (x yz))

y^{^{''}} + 45 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 0

n = 1 \sum \infty n e^{3 n}

What is the integral from 0 to 1 of 19 1/(x^p) ?
The integral from 0 to 1 of 19 1/(x^p) is (19)/(-p+1)