What is the integral from 0 to 1 of 6(1+sqrt(x))^4 ?

The integral from 0 to 1 of 6(1+sqrt(x))^4 is 258/5

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 6(1+sqrt(x))^4

\int_{0}^{1} 6 (1 + x)^{4} d x

\frac{258}{5}

Solution steps

\int_{0}^{1} 6 (1 + x)^{4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{0}^{1} (1 + x)^{4} d x

Expand

(1 + x)^{4} : 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{2}

= 6 \cdot \int_{0}^{1} 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{2} 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

= 6 (\int_{0}^{1} 1 d x + \int_{0}^{1} 4 x d x + \int_{0}^{1} 6 x d x + \int_{0}^{1} 4 x^{\frac{3}{2}} d x + \int_{0}^{1} x^{2} d x)

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

\int_{0}^{1} 4 x d x = \frac{8}{3}

\int_{0}^{1} 6 x d x = 3

\int_{0}^{1} 4 x^{\frac{3}{2}} d x = \frac{8}{5}

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

= 6 (1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3})

Simplify

6 (1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3}) : \frac{258}{5}

= \frac{258}{5}

\int \frac{1}{x + 3 - 1} d x

\int sin (x) + C d x

\int e^{8 + x} d x

\int_{1}^{3} (2 x^{2} - 3 x + 5) d x

e^{13 t} sin (t)

What is the integral from 0 to 1 of 6(1+sqrt(x))^4 ?
The integral from 0 to 1 of 6(1+sqrt(x))^4 is 258/5