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

The integral from 1 to 2 of xsqrt(2-x) is 14/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{14}{15}

Decimal

0.93333 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - (- u + 2) u d u

= \int_{1}^{0} - (- u + 2) u d u

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

= - \int_{0}^{1} - (- u + 2) u d u

Take the constant out:

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

= - (- \int_{0}^{1} (- u + 2) u d u)

Expand

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

= - (- \int_{0}^{1} - u^{\frac{3}{2}} + 2 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_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} 2 u d u))

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

\int_{0}^{1} 2 u d u = \frac{4}{3}

= - (- (- \frac{2}{5} + \frac{4}{3}))

Simplify

- (- (- \frac{2}{5} + \frac{4}{3})) : \frac{14}{15}

= \frac{14}{15}

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

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

\int_{0}^{4} 1000 x d x

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

\int_{1}^{5} 6 x^{2} d x

What is the integral from 1 to 2 of xsqrt(2-x) ?
The integral from 1 to 2 of xsqrt(2-x) is 14/15