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

The integral from 0 to 1 of x^2sqrt(2+x) is 8sqrt(3)+(-128sqrt(2)-702sqrt(3))/(105)

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 x^2sqrt(2+x)

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

83 + \frac{- 128 2 - 702 3}{105}

Solution steps

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

Apply u-substitution

: \int_{2}^{3} (u - 2)^{2} u d u

= \int_{2}^{3} (u - 2)^{2} u d u

Expand

(u - 2)^{2} u : u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 u

= \int_{2}^{3} u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{3} u^{\frac{5}{2}} d u - \int_{2}^{3} 4 u^{\frac{3}{2}} d u + \int_{2}^{3} 4 u d u

\int_{2}^{3} u^{\frac{5}{2}} d u = \frac{54 3 - 16 2}{7}

\int_{2}^{3} 4 u^{\frac{3}{2}} d u = \frac{4 ( 18 3 - 8 2 )}{5}

\int_{2}^{3} 4 u d u = 4 (23 - \frac{4 2}{3})

= \frac{54 3 - 16 2}{7} - \frac{4 ( 18 3 - 8 2 )}{5} + 4 (23 - \frac{4 2}{3})

Simplify

\frac{54 3 - 16 2}{7} - \frac{4 ( 18 3 - 8 2 )}{5} + 4 (23 - \frac{4 2}{3}) : 83 + \frac{- 128 2 - 702 3}{105}

= 83 + \frac{- 128 2 - 702 3}{105}

\int_{1}^{2} π (x^{- 1251})^{2} d x

\int_{1}^{4} \frac{1}{2 y} d y

\int_{0}^{4} 3 x (4 - x) d x

\int_{0}^{a} e^{- s t} d t

\int_{1}^{3} \frac{1}{x ^{2} - 1} d x

What is the integral from 0 to 1 of x^2sqrt(2+x) ?
The integral from 0 to 1 of x^2sqrt(2+x) is 8sqrt(3)+(-128sqrt(2)-702sqrt(3))/(105)