What is the integral from 4 to 5 of x^2sqrt(x-4) ?

The integral from 4 to 5 of x^2sqrt(x-4) is 1486/105

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 5 of x^2sqrt(x-4)

\int_{4}^{5} x^{2} x - 4 d x

\frac{1486}{105}

Decimal Notation

14.15238 \dots

Solution steps

\int_{4}^{5} x^{2} x - 4 d x

Apply u-substitution

: \int_{0}^{1} (u + 4)^{2} u d u

= \int_{0}^{1} (u + 4)^{2} u d u

Expand

(u + 4)^{2} u : u^{\frac{5}{2}} + 8 u^{\frac{3}{2}} + 16 u

= \int_{0}^{1} u^{\frac{5}{2}} + 8 u^{\frac{3}{2}} + 16 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{5}{2}} d u + \int_{0}^{1} 8 u^{\frac{3}{2}} d u + \int_{0}^{1} 16 u d u

\int_{0}^{1} u^{\frac{5}{2}} d u = \frac{2}{7}

\int_{0}^{1} 8 u^{\frac{3}{2}} d u = \frac{16}{5}

\int_{0}^{1} 16 u d u = \frac{32}{3}

= \frac{2}{7} + \frac{16}{5} + \frac{32}{3}

Simplify

\frac{2}{7} + \frac{16}{5} + \frac{32}{3} : \frac{1486}{105}

= \frac{1486}{105}

\frac{\partial}{\partial x} (\frac{20}{8 + x ^{2} + y ^{2}})

\frac{\partial}{\partial t} (\frac{8.31 t}{V ( t )})

x \to 2 lim (\frac{2 x - 6}{x - 1})

t e^{8 t}

x \to 9 - lim (\frac{5 x}{x - 9})

What is the integral from 4 to 5 of x^2sqrt(x-4) ?
The integral from 4 to 5 of x^2sqrt(x-4) is 1486/105