What is the integral of x^2sqrt(6-x) ?

The integral of x^2sqrt(6-x) is -2/7 (6-x)^{7/2}+24/5 (6-x)^{5/2}-24(6-x)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2sqrt(6-x)

\int x^{2} 6 - x d x

- \frac{2}{7} (6 - x)^{\frac{7}{2}} + \frac{24}{5} (6 - x)^{\frac{5}{2}} - 24 (6 - x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 6 - x d x

Apply u-substitution

: \int - (- u + 6)^{2} u d u

= \int - (- u + 6)^{2} u d u

Take the constant out:

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

= - \int (- u + 6)^{2} u d u

Expand

(- u + 6)^{2} u : u^{\frac{5}{2}} - 12 u^{\frac{3}{2}} + 36 u

= - \int u^{\frac{5}{2}} - 12 u^{\frac{3}{2}} + 36 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

= - (\int u^{\frac{5}{2}} d u - \int 12 u^{\frac{3}{2}} d u + \int 36 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 12 u^{\frac{3}{2}} d u = \frac{24}{5} u^{\frac{5}{2}}

\int 36 u d u = 24 u^{\frac{3}{2}}

= - (\frac{2}{7} u^{\frac{7}{2}} - \frac{24}{5} u^{\frac{5}{2}} + 24 u^{\frac{3}{2}})

Substitute back

u = 6 - x

= - (\frac{2}{7} (6 - x)^{\frac{7}{2}} - \frac{24}{5} (6 - x)^{\frac{5}{2}} + 24 (6 - x)^{\frac{3}{2}})

Simplify

= - \frac{2}{7} (6 - x)^{\frac{7}{2}} + \frac{24}{5} (6 - x)^{\frac{5}{2}} - 24 (6 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{7} (6 - x)^{\frac{7}{2}} + \frac{24}{5} (6 - x)^{\frac{5}{2}} - 24 (6 - x)^{\frac{3}{2}} + C

\int (x^{4} - \frac{6}{x ^{4}}) d x

\int \frac{x}{( x ^{2} )} d x

\int \frac{- 15}{x ^{2} + 1} d x

\int 3 x^{2} y^{2} z d x

\int \frac{cos ( x )}{cos ( x ) + 1} d x

What is the integral of x^2sqrt(6-x) ?
The integral of x^2sqrt(6-x) is -2/7 (6-x)^{7/2}+24/5 (6-x)^{5/2}-24(6-x)^{3/2}+C