What is the integral from 0 to 8 of (3x)/(sqrt(1+x)) ?

The integral from 0 to 8 of (3x)/(sqrt(1+x)) is 40

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 8 of (3x)/(sqrt(1+x))

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

40

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{3} 2 (u^{2} - 1) d u

= 3 \cdot \int_{1}^{3} 2 (u^{2} - 1) d u

Take the constant out:

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

= 3 \cdot 2 \cdot \int_{1}^{3} u^{2} - 1 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

= 3 \cdot 2 (\int_{1}^{3} u^{2} d u - \int_{1}^{3} 1 d u)

\int_{1}^{3} u^{2} d u = \frac{26}{3}

\int_{1}^{3} 1 d u = 2

= 3 \cdot 2 (\frac{26}{3} - 2)

Simplify

3 \cdot 2 (\frac{26}{3} - 2) : 40

= 40

\int_{0}^{2 π} 3 sin (x) d x

\int_{1}^{5} ∣ 4 x - 8 ∣ d x

\int_{x}^{6} sin (t^{4}) d t

\int_{4}^{9} (1 + x) x d x

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

What is the integral from 0 to 8 of (3x)/(sqrt(1+x)) ?
The integral from 0 to 8 of (3x)/(sqrt(1+x)) is 40