What is the integral from 4 to 7 of (2x)/(sqrt(8-x)) ?

The integral from 4 to 7 of (2x)/(sqrt(8-x)) is 68/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 7 of (2x)/(sqrt(8-x))

\int_{4}^{7} \frac{2 x}{8 - x} d x

\frac{68}{3}

Decimal

22.66666 \dots

Solution steps

\int_{4}^{7} \frac{2 x}{8 - x} d x

Take the constant out:

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

= 2 \cdot \int_{4}^{7} \frac{x}{8 - x} d x

Apply u-substitution

: \int_{2}^{1} - 2 (- u^{2} + 8) d u

= 2 \cdot \int_{2}^{1} - 2 (- u^{2} + 8) d u

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

= 2 (- \int_{1}^{2} - 2 (- u^{2} + 8) d u)

Take the constant out:

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

= 2 (- (- 2 \cdot \int_{1}^{2} - u^{2} + 8 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 (- (- 2 (- \int_{1}^{2} u^{2} d u + \int_{1}^{2} 8 d u)))

\int_{1}^{2} u^{2} d u = \frac{7}{3}

\int_{1}^{2} 8 d u = 8

= 2 (- (- 2 (- \frac{7}{3} + 8)))

Simplify

2 (- (- 2 (- \frac{7}{3} + 8))) : \frac{68}{3}

= \frac{68}{3}

\int_{x}^{0} e^{- 10 t} d t

\int_{0}^{1} cos (w x) d x

\int_{- 1}^{5} 2 x + 6 - x + 1 d x

\int_{- 2}^{3} (9 - x^{2}) - (3 - x) d x

\int_{0}^{x^{3}} e^{- 2 t} d t

What is the integral from 4 to 7 of (2x)/(sqrt(8-x)) ?
The integral from 4 to 7 of (2x)/(sqrt(8-x)) is 68/3