What is the integral from 0 to 1 of (2x^2+sqrt(x)) ?

The integral from 0 to 1 of (2x^2+sqrt(x)) is 4/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of (2x^2+sqrt(x))

\int_{0}^{1} (2 x^{2} + x) d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{0}^{1} 2 x^{2} + x 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} 2 x^{2} d x + \int_{0}^{1} x d x

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

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

= \frac{2}{3} + \frac{2}{3}

Simplify

\frac{2}{3} + \frac{2}{3} : \frac{4}{3}

= \frac{4}{3}

What is the integral from 0 to 1 of (2x^2+sqrt(x)) ?
The integral from 0 to 1 of (2x^2+sqrt(x)) is 4/3