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

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

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 sqrt(x)+x^2

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

1

Solution steps

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

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

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

= \frac{2}{3} + \frac{1}{3}

Simplify

\frac{2}{3} + \frac{1}{3} : 1

= 1

\int_{- 1}^{1} e^{x} x d x

\int_{- 1}^{1} x (x + 4)^{2} d x

\int_{0}^{\frac{π}{2}} sec (x) tan (x) d x

\int_{1}^{4} ln (2 x) d x

\int_{- 1}^{4} 2 x d x

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