What is the integral from 0 to 3 of (3-x)^{(-1)/2} ?

The integral from 0 to 3 of (3-x)^{(-1)/2} is 2*3^{1/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of (3-x)^{(-1)/2}

\int_{0}^{3} (3 - x)^{\frac{- 1}{2}} d x

2 \cdot 3^{\frac{1}{2}}

Decimal

3.46410 \dots

Solution steps

\int_{0}^{3} (3 - x)^{\frac{- 1}{2}} d x

Simplify

(3 - x)^{\frac{- 1}{2}} : \frac{1}{( 3 - x ) ^{\frac{1}{2}}}

= \int_{0}^{3} \frac{1}{( 3 - x ) ^{\frac{1}{2}}} d x

Apply Integration By Parts

: [\frac{x}{( 3 - x ) ^{\frac{1}{2}}} - \int \frac{x}{2 ( 3 - x ) ^{\frac{3}{2}}} d x]_{0}^{3}

= [\frac{x}{( 3 - x ) ^{\frac{1}{2}}} - \int \frac{x}{2 ( 3 - x ) ^{\frac{3}{2}}} d x]_{0}^{3}

\int \frac{x}{2 ( 3 - x ) ^{\frac{3}{2}}} d x = - \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}}

= [\frac{x}{( 3 - x ) ^{\frac{1}{2}}} - (- \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}})]_{0}^{3}

Simplify

[\frac{x}{( 3 - x ) ^{\frac{1}{2}}} - (- \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}})]_{0}^{3} : [- 2 (- x + 3)^{\frac{1}{2}}]_{0}^{3}

= [- 2 (- x + 3)^{\frac{1}{2}}]_{0}^{3}

Compute the boundaries:

2 \cdot 3^{\frac{1}{2}}

= 2 \cdot 3^{\frac{1}{2}}

\int_{- 6}^{1} x^{3} d x

\int_{1}^{2} (e^{3 t} - \frac{4}{t}) d t

\int_{0}^{1} e^{x} - e^{7 x - 6} d x

\int_{3}^{5} x e^{x^{2}} d x

\int_{1}^{3} (\frac{3 x ^{2} + 4 x + 1}{x}) d x

What is the integral from 0 to 3 of (3-x)^{(-1)/2} ?
The integral from 0 to 3 of (3-x)^{(-1)/2} is 2*3^{1/2}