What is the integral from 0 to 1 of 7 1/(x^p) ?

The integral from 0 to 1 of 7 1/(x^p) is 7/(-p+1)

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integral from 0 to 1 of 7 1/(x^p)

\int_{0}^{1} 7 \frac{1}{x ^{p}} d x

\frac{7}{- p + 1}

Solution steps

\int_{0}^{1} 7 \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

= 7 \cdot \int_{0}^{1} \frac{1}{x ^{p}} d x

Apply the Power Rule

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

= 7 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{- p + 1}

= 7 \cdot \frac{1}{- p + 1}

Simplify

= \frac{7}{- p + 1}

x \to 3 lim (x^{3})

\int \frac{x + 2}{2 x + 2} d x

x \to 2 - lim (\frac{9 x}{x - 2})

\frac{\partial}{\partial y} (4 x + 9 y - 360)

\frac{( 2 x + 1 ) ( x ^{3} + 3 )}{x}

What is the integral from 0 to 1 of 7 1/(x^p) ?
The integral from 0 to 1 of 7 1/(x^p) is 7/(-p+1)