What is the integral from 0 to 1 of 40 1/(x^p) ?

The integral from 0 to 1 of 40 1/(x^p) is (40)/(-p+1)

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integral from 0 to 1 of 40 1/(x^p)

\int_{0}^{1} 40 \frac{1}{x ^{p}} d x

\frac{40}{- p + 1}

Solution steps

\int_{0}^{1} 40 \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

= 40 \cdot \int_{0}^{1} \frac{1}{x ^{p}} d x

Apply the Power Rule

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

= 40 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{- p + 1}

= 40 \cdot \frac{1}{- p + 1}

Simplify

= \frac{40}{- p + 1}

\int \frac{( ( ( x ) + 1 ) \cdot ( x - 1 ) )}{( x )} d x

f (x) = x^{2} + 2, (- 2, 6)

\int 2^{x - 1} d x

(2) (e^{3 x x})

What is the integral from 0 to 1 of 40 1/(x^p) ?
The integral from 0 to 1 of 40 1/(x^p) is (40)/(-p+1)