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

The integral from 1 to 2 of (x+sqrt(x))^2 is 23/6+(2(8sqrt(2)-2))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{23}{6} + \frac{2 ( 8 2 - 2 )}{5}

Decimal Notation

7.55881 \dots

Solution steps

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

Expand

(x + x)^{2} : x^{2} + 2 x^{\frac{3}{2}} + x

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

\int_{1}^{2} 2 x^{\frac{3}{2}} d x = \frac{2 ( 8 2 - 2 )}{5}

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

= \frac{7}{3} + \frac{2 ( 8 2 - 2 )}{5} + \frac{3}{2}

Combine the fractions

\frac{7}{3} + \frac{3}{2} : \frac{23}{6}

= \frac{23}{6} + \frac{2 ( 8 2 - 2 )}{5}

\int_{0}^{1} 2 x e^{3 x} d x

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

\int_{0}^{2} x^{2} - 5 d x

\int_{- 1}^{1} 2 y^{2} d y

\int_{0}^{y^{2}} e^{- x} d x

What is the integral from 1 to 2 of (x+sqrt(x))^2 ?
The integral from 1 to 2 of (x+sqrt(x))^2 is 23/6+(2(8sqrt(2)-2))/5