What is the integral from 1 to 2 of 4/((x-1)^{1/4)} ?

The integral from 1 to 2 of 4/((x-1)^{1/4)} is 16/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of 4/((x-1)^{1/4)}

\int_{1}^{2} \frac{4}{( x - 1 ) ^{\frac{1}{4}}} d x

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

\int_{1}^{2} \frac{4}{( x - 1 ) ^{\frac{1}{4}}} d x

Take the constant out:

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

= 4 \cdot \int_{1}^{2} \frac{1}{( x - 1 ) ^{\frac{1}{4}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{( x - 1 ) ^{\frac{1}{4}}} = (x - 1)^{- \frac{1}{4}}

= 4 \cdot \int_{1}^{2} (x - 1)^{- \frac{1}{4}} d x

Apply u-substitution

: \int_{0}^{1} \frac{1}{u ^{\frac{1}{4}}} d u

= 4 \cdot \int_{0}^{1} \frac{1}{u ^{\frac{1}{4}}} d u

Apply the Power Rule

: [\frac{4}{3} u^{\frac{3}{4}}]_{0}^{1}

= 4 [\frac{4}{3} u^{\frac{3}{4}}]_{0}^{1}

Compute the boundaries:

\frac{4}{3}

= 4 \cdot \frac{4}{3}

Simplify

= \frac{16}{3}

\frac{d}{d x} (\frac{1}{( - x ^{2} + 3 x + 3 ) ^{4}})

x \to 3 + lim (\frac{3 x}{x - 3})

x \to 1 lim (\frac{x ^{m} - 1}{x - 1})

f (x) = \frac{x ^{2}}{x - 3}, at x = 0

\frac{\partial}{\partial x} (cos (x + 9 y))

What is the integral from 1 to 2 of 4/((x-1)^{1/4)} ?
The integral from 1 to 2 of 4/((x-1)^{1/4)} is 16/3