What is the integral from 0 to 1 of (37)/(x^p) ?

The integral from 0 to 1 of (37)/(x^p) is (37)/(-p+1)

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integral from 0 to 1 of (37)/(x^p)

\int_{0}^{1} \frac{37}{x ^{p}} d x

\frac{37}{- p + 1}

Solution steps

\int_{0}^{1} \frac{37}{x ^{p}} d x

Take the constant out:

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

= 37 \cdot \int_{0}^{1} \frac{1}{x ^{p}} d x

Apply the Power Rule

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

= 37 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{- p + 1}

= 37 \cdot \frac{1}{- p + 1}

Simplify

= \frac{37}{- p + 1}

\frac{d}{d x} (e^{3 x^{2}} \cdot 6 x)

y = 2 x^{3} + 3 x^{2} - 12 x + 6

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + y = 1

x \to 2 lim (- 10 x + (10))

f (x) = 28 x - 0.01 x^{2}

What is the integral from 0 to 1 of (37)/(x^p) ?
The integral from 0 to 1 of (37)/(x^p) is (37)/(-p+1)