What is the integral from 0 to 4 of 1/((9-3x)^{1/3)} ?

The integral from 0 to 4 of 1/((9-3x)^{1/3)} is (9^{2/3}-3^{2/3})/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of 1/((9-3x)^{1/3)}

\int_{0}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

\frac{9 ^{\frac{2}{3}} - 3 ^{\frac{2}{3}}}{2}

Decimal Notation

1.12333 \dots

Solution steps

\int_{0}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

Find undefined (singularity) points:

x = 3

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{3} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x + \int_{3}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

\int_{0}^{3} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x = \frac{9 ^{\frac{2}{3}}}{2}

\int_{3}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x = - \frac{( - 1 ) ^{\frac{2}{3}} \cdot 3 ^{\frac{2}{3}}}{2}

= \frac{9 ^{\frac{2}{3}}}{2} - \frac{( - 1 ) ^{\frac{2}{3}} \cdot 3 ^{\frac{2}{3}}}{2}

Simplify

\frac{9 ^{\frac{2}{3}}}{2} - \frac{( - 1 ) ^{\frac{2}{3}} \cdot 3 ^{\frac{2}{3}}}{2} : \frac{9 ^{\frac{2}{3}} - 3 ^{\frac{2}{3}}}{2}

= \frac{9 ^{\frac{2}{3}} - 3 ^{\frac{2}{3}}}{2}

\int 3 x^{2} + C d x

f (x) = 3 x + \frac{5}{x}

x \to 3 + lim (x x - 3)

x y^{^{'}} + 3 y = \frac{4 e ^{- 3 x}}{x ^{2}}

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

What is the integral from 0 to 4 of 1/((9-3x)^{1/3)} ?
The integral from 0 to 4 of 1/((9-3x)^{1/3)} is (9^{2/3}-3^{2/3})/2