What is the integral from 0 to 2 of (2x)/(3-x) ?

The integral from 0 to 2 of (2x)/(3-x) is 2(-2+3ln(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of (2x)/(3-x)

\int_{0}^{2} \frac{2 x}{3 - x} d x

2 (- 2 + 3 ln (3))

Decimal

2.59167 \dots

Solution steps

\int_{0}^{2} \frac{2 x}{3 - x} d x

Take the constant out:

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

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

Apply u-substitution

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Expand

\frac{- u + 3}{u} : - 1 + \frac{3}{u}

= 2 (- (- \int_{1}^{3} - 1 + \frac{3}{u} 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

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

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

\int_{1}^{3} \frac{3}{u} d u = 3 ln (3)

= 2 (- (- (- 2 + 3 ln (3))))

Simplify

= 2 (- 2 + 3 ln (3))

What is the integral from 0 to 2 of (2x)/(3-x) ?
The integral from 0 to 2 of (2x)/(3-x) is 2(-2+3ln(3))