What is the integral from 0 to 4 of sqrt(x)(x+2) ?

The integral from 0 to 4 of sqrt(x)(x+2) is 352/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of sqrt(x)(x+2)

\int_{0}^{4} x (x + 2) d x

\frac{352}{15}

Decimal Notation

23.46666 \dots

Solution steps

\int_{0}^{4} x (x + 2) d x

Expand

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

= \int_{0}^{4} x^{\frac{3}{2}} + 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_{0}^{4} x^{\frac{3}{2}} d x + \int_{0}^{4} 2 x d x

\int_{0}^{4} x^{\frac{3}{2}} d x = \frac{64}{5}

\int_{0}^{4} 2 x d x = \frac{32}{3}

= \frac{64}{5} + \frac{32}{3}

Simplify

\frac{64}{5} + \frac{32}{3} : \frac{352}{15}

= \frac{352}{15}

x^{^{''}} + 5 x^{^{'}} = 7

\int \frac{4}{t + 1} d t

\frac{d}{d x} (x^{6} e^{8 x})

\int \frac{x + 2}{x ^{2} - 2 x + 10} d x

ln (x^{5})

What is the integral from 0 to 4 of sqrt(x)(x+2) ?
The integral from 0 to 4 of sqrt(x)(x+2) is 352/15