What is the integral from 0 to 1 of (42)/(x^p) ?

The integral from 0 to 1 of (42)/(x^p) is (42)/(-p+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (42)/(x^p)

\int_{0}^{1} \frac{42}{x ^{p}} d x

\frac{42}{- p + 1}

Solution steps

\int_{0}^{1} \frac{42}{x ^{p}} d x

Take the constant out:

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

= 42 \cdot \int_{0}^{1} \frac{1}{x ^{p}} d x

Apply the Power Rule

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

= 42 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{- p + 1}

= 42 \cdot \frac{1}{- p + 1}

Simplify

= \frac{42}{- p + 1}

\int_{0}^{2} x 11 - x^{2} d x

\int (x^{2} + 5) e^{- x} d x

\int \frac{( x ^{5} + 2 x ^{2} - 1 )}{x ^{4}} d x

2 y^{^{'}} + y = \frac{3}{2} t^{2}

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

What is the integral from 0 to 1 of (42)/(x^p) ?
The integral from 0 to 1 of (42)/(x^p) is (42)/(-p+1)