What is the integral from 0 to 1 of 3(1+sqrt(x))^5 ?

The integral from 0 to 1 of 3(1+sqrt(x))^5 is 321/7

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 3(1+sqrt(x))^5

\int_{0}^{1} 3 (1 + x)^{5} d x

\frac{321}{7}

Decimal Notation

45.85714 \dots

Solution steps

\int_{0}^{1} 3 (1 + x)^{5} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} (1 + x)^{5} d x

Expand

(1 + x)^{5} : 1 + 5 x + 10 x + 10 x^{\frac{3}{2}} + 5 x^{2} + x^{\frac{5}{2}}

= 3 \cdot \int_{0}^{1} 1 + 5 x + 10 x + 10 x^{\frac{3}{2}} + 5 x^{2} + x^{\frac{5}{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

= 3 (\int_{0}^{1} 1 d x + \int_{0}^{1} 5 x d x + \int_{0}^{1} 10 x d x + \int_{0}^{1} 10 x^{\frac{3}{2}} d x + \int_{0}^{1} 5 x^{2} d x + \int_{0}^{1} x^{\frac{5}{2}} d x)

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

\int_{0}^{1} 5 x d x = \frac{10}{3}

\int_{0}^{1} 10 x d x = 5

\int_{0}^{1} 10 x^{\frac{3}{2}} d x = 4

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

\int_{0}^{1} x^{\frac{5}{2}} d x = \frac{2}{7}

= 3 (1 + \frac{10}{3} + 5 + 4 + \frac{5}{3} + \frac{2}{7})

Simplify

3 (1 + \frac{10}{3} + 5 + 4 + \frac{5}{3} + \frac{2}{7}) : \frac{321}{7}

= \frac{321}{7}

\int_{0}^{1} (2 x - 3) d x

\frac{d y}{d x} = - 10 y

f (x) = \frac{x ^{3} + 2 x + 7}{x ^{2} - 5}

\int_{1}^{1.5} \frac{2}{5} (x + 1) d x

f (t) = 6 t^{3} - 54 t^{2} + 70

What is the integral from 0 to 1 of 3(1+sqrt(x))^5 ?
The integral from 0 to 1 of 3(1+sqrt(x))^5 is 321/7