What is the integral from 2 to 3 of (19)/(sqrt(3-x)) ?

The integral from 2 to 3 of (19)/(sqrt(3-x)) is 38

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 2 to 3 of (19)/(sqrt(3-x))

\int_{2}^{3} \frac{19}{3 - x} d x

38

Solution steps

\int_{2}^{3} \frac{19}{3 - x} d x

Take the constant out:

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

= 19 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

= 19 \cdot \int_{1}^{0} - \frac{1}{u} d u

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

= 19 (- \int_{0}^{1} - \frac{1}{u} d u)

Take the constant out:

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

= 19 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

= 19 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

19 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 19 [2 u]_{0}^{1}

= 19 [2 u]_{0}^{1}

Compute the boundaries:

2

= 19 \cdot 2

Simplify

= 38

What is the integral from 2 to 3 of (19)/(sqrt(3-x)) ?
The integral from 2 to 3 of (19)/(sqrt(3-x)) is 38