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

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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

- 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}

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