What is the integral from 0 to 1 of (4x+sqrt(x)) ?

The integral from 0 to 1 of (4x+sqrt(x)) is 8/3

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+sqrt(x))

\int_{0}^{1} (4 x + x) d x

\frac{8}{3}

Solution steps

\int_{0}^{1} 4 x + 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} 4 x d x + \int_{0}^{1} x d x

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

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

= 2 + \frac{2}{3}

Simplify

2 + \frac{2}{3} : \frac{8}{3}

= \frac{8}{3}

\int_{1}^{e^{2}} π (ln (x))^{2} d x

\int_{0}^{3} \frac{16}{3} d x

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

\int_{0}^{1} ln (y + 1) d y

\int_{- 3}^{2} 4 sin (x^{2}) - 1 d x

What is the integral from 0 to 1 of (4x+sqrt(x)) ?
The integral from 0 to 1 of (4x+sqrt(x)) is 8/3