What is the integral from 0 to 9 of x/(sqrt(4+5x)) ?

The integral from 0 to 9 of x/(sqrt(4+5x)) is 22/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 9 of x/(sqrt(4+5x))

\int_{0}^{9} \frac{x}{4 + 5 x} d x

\frac{22}{3}

Decimal

7.33333 \dots

Solution steps

\int_{0}^{9} \frac{x}{4 + 5 x} d x

Apply u-substitution

: \int_{2}^{7} \frac{2 ( u ^{2} - 4 )}{25} d u

= \int_{2}^{7} \frac{2 ( u ^{2} - 4 )}{25} d u

Take the constant out:

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

= \frac{2}{25} \cdot \int_{2}^{7} u^{2} - 4 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

= \frac{2}{25} (\int_{2}^{7} u^{2} d u - \int_{2}^{7} 4 d u)

\int_{2}^{7} u^{2} d u = \frac{335}{3}

\int_{2}^{7} 4 d u = 20

= \frac{2}{25} (\frac{335}{3} - 20)

Simplify

\frac{2}{25} (\frac{335}{3} - 20) : \frac{22}{3}

= \frac{22}{3}

\int_{0}^{5} (x - 2)^{3} d x

\int_{- 2}^{3} (3 - ∣ x ∣) d x

\int_{5}^{9} x d x

\int_{3}^{5} \frac{x ^{2} + x + 6}{x - 2} d x

\int_{5}^{10} x^{2} d x

What is the integral from 0 to 9 of x/(sqrt(4+5x)) ?
The integral from 0 to 9 of x/(sqrt(4+5x)) is 22/3