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

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

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 sqrt(2x)

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

- \frac{2}{3} + 2

Decimal

0.94280 \dots

Solution steps

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

Apply Integration By Parts

: [2 x x - \int \frac{x}{2} d x]_{0}^{1}

= [2 x x - \int \frac{x}{2} d x]_{0}^{1}

\int \frac{x}{2} d x = \frac{2}{3} x^{\frac{3}{2}}

= [2 x x - \frac{2}{3} x^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

- \frac{2}{3} + 2

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

\int_{3}^{5} 6 x^{2} (x^{3} + 4) d x

\int_{- 1}^{1} - 2 cos (\frac{π x}{3}) d x

\int_{1}^{3} 4 - x d x

\int_{0}^{y} e^{- x} d x

\int_{0}^{6} \frac{5}{12} x^{2} - 5 x d x

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