What is the integral from 4 to 6 of x^2sqrt(x-4) ?

The integral from 4 to 6 of x^2sqrt(x-4) is (3824sqrt(2))/(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 6 of x^2sqrt(x-4)

\int_{4}^{6} x^{2} x - 4 d x

\frac{3824 2}{105}

Decimal Notation

51.50431 \dots

Solution steps

\int_{4}^{6} x^{2} x - 4 d x

Apply u-substitution

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

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

Expand

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

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

\int_{0}^{2} u^{\frac{5}{2}} d u = \frac{16 2}{7}

\int_{0}^{2} 8 u^{\frac{3}{2}} d u = \frac{64 2}{5}

\int_{0}^{2} 16 u d u = \frac{64 2}{3}

= \frac{16 2}{7} + \frac{64 2}{5} + \frac{64 2}{3}

Simplify

\frac{16 2}{7} + \frac{64 2}{5} + \frac{64 2}{3} : \frac{3824 2}{105}

= \frac{3824 2}{105}

\int_{2}^{4} (x y^{2} + x^{2} + y + 4) d y

\int_{- 5}^{- 3} x^{2} - 4 d x

\int_{2}^{9} \frac{e ^{x}}{x} d x

\int_{0}^{4} x (3 - 5 x) d x

\int_{0}^{8} 6 x ln (x) d x

What is the integral from 4 to 6 of x^2sqrt(x-4) ?
The integral from 4 to 6 of x^2sqrt(x-4) is (3824sqrt(2))/(105)