What is the integral of (sqrt(3-x))(x) ?

The integral of (sqrt(3-x))(x) is 2/5 (3-x)^{5/2}-2(3-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 (sqrt(3-x))(x)

\int (3 - x) (x) d x

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

Solution steps

\int 3 - x x d x

Apply u-substitution

: \int - (- u + 3) u d u

= \int - (- u + 3) u d u

Take the constant out:

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

= - \int (- u + 3) u d u

Expand

(- u + 3) u : - u^{\frac{3}{2}} + 3 u

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

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

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

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

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

Substitute back

u = 3 - x

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

Simplify

= \frac{2}{5} (3 - x)^{\frac{5}{2}} - 2 (3 - x)^{\frac{3}{2}}

Add a constant to the solution

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

\int x x + 14 d x

\frac{d}{d x} (m x^{m + 5} + 6 x^{m} y^{n} + n x^{n + 3})

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

\int sec (2 θ) d θ

f (x) = 4 x^{2} - 2 x + 5

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