What is the integral from 0 to 1 of xsqrt(3-2x) ?

The integral from 0 to 1 of xsqrt(3-2x) is -(2-3sqrt(3))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of xsqrt(3-2x)

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

- \frac{2 - 3 3}{5}

Solution steps

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

Apply u-substitution

: \int_{3}^{1} \frac{u ( u - 3 )}{4} d u

= \int_{3}^{1} \frac{u ( u - 3 )}{4} d u

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

= - \int_{1}^{3} \frac{u ( u - 3 )}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{1}^{3} u (u - 3) d u

Expand

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

= - \frac{1}{4} \cdot \int_{1}^{3} 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

= - \frac{1}{4} (\int_{1}^{3} u^{\frac{3}{2}} d u - \int_{1}^{3} 3 u d u)

\int_{1}^{3} u^{\frac{3}{2}} d u = \frac{18 3 - 2}{5}

\int_{1}^{3} 3 u d u = 63 - 2

= - \frac{1}{4} (\frac{18 3 - 2}{5} - (63 - 2))

Simplify

- \frac{1}{4} (\frac{18 3 - 2}{5} - (63 - 2)) : - \frac{2 - 3 3}{5}

= - \frac{2 - 3 3}{5}

\int_{1}^{2} t^{2} d t

\int_{- π}^{π} sin (n x) sin (m x) d x

\int_{0}^{2 π} cos (θ) sin^{2} (θ) d θ

\int_{0}^{2} (x + 2)^{3} d x

\int_{0}^{4} (x - \frac{x}{2}) d x

What is the integral from 0 to 1 of xsqrt(3-2x) ?
The integral from 0 to 1 of xsqrt(3-2x) is -(2-3sqrt(3))/5