What is the integral from-3 to 1 of 1/((2x+6)^{2/3)} ?

The integral from-3 to 1 of 1/((2x+6)^{2/3)} is 3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-3 to 1 of 1/((2x+6)^{2/3)}

\int_{- 3}^{1} \frac{1}{( 2 x + 6 ) ^{\frac{2}{3}}} d x

3

Solution steps

\int_{- 3}^{1} \frac{1}{( 2 x + 6 ) ^{\frac{2}{3}}} d x

Apply u-substitution

= \int_{0}^{8} \frac{1}{2 u ^{\frac{2}{3}}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{8} \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

= \frac{1}{2} [3 u^{\frac{1}{3}}]_{0}^{8}

Compute the boundaries:

6

= \frac{1}{2} \cdot 6

Simplify

= 3

What is the integral from-3 to 1 of 1/((2x+6)^{2/3)} ?
The integral from-3 to 1 of 1/((2x+6)^{2/3)} is 3