What is the integral from 0 to 1 of 4x-6(x^2)^{1/3} ?

The integral from 0 to 1 of 4x-6(x^2)^{1/3} is -8/5

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 4x-6(x^2)^{1/3}

\int_{0}^{1} 4 x - 6 (x^{2})^{\frac{1}{3}} d x

- \frac{8}{5}

Decimal Notation

- 1.6

Solution steps

\int_{0}^{1} 4 x - 6 (x^{2})^{\frac{1}{3}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 4 x d x - \int_{0}^{1} 6 (x^{2})^{\frac{1}{3}} d x

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} 6 (x^{2})^{\frac{1}{3}} d x = \frac{18}{5}

= 2 - \frac{18}{5}

Simplify

2 - \frac{18}{5} : - \frac{8}{5}

= - \frac{8}{5}

\int_{- 1}^{3} (3 x - 5) (x + 2) d x

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

\int_{2}^{4} 8 - 2 x d x

\int_{0}^{3} x^{2} - 2 x - 3 d x

\int_{0}^{0.1} 600 x d x

What is the integral from 0 to 1 of 4x-6(x^2)^{1/3} ?
The integral from 0 to 1 of 4x-6(x^2)^{1/3} is -8/5