What is the integral from 0 to 5 of x^3sqrt(x^2+25) ?

The integral from 0 to 5 of x^3sqrt(x^2+25) is 2500sqrt(2)-625-(25(250sqrt(2)-125))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 5 of x^3sqrt(x^2+25)

\int_{0}^{5} x^{3} x^{2} + 25 d x

25002 - 625 - \frac{25 ( 250 2 - 125 )}{3}

Decimal Notation

1005.92231 \dots

Solution steps

\int_{0}^{5} x^{3} x^{2} + 25 d x

Apply u-substitution

: \int_{5}^{52} (u^{2} - 25) u^{2} d u

= \int_{5}^{52} (u^{2} - 25) u^{2} d u

Expand

(u^{2} - 25) u^{2} : u^{4} - 25 u^{2}

= \int_{5}^{52} u^{4} - 25 u^{2} 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_{5}^{52} u^{4} d u - \int_{5}^{52} 25 u^{2} d u

\int_{5}^{52} u^{4} d u = 25002 - 625

\int_{5}^{52} 25 u^{2} d u = \frac{25 ( 250 2 - 125 )}{3}

= 25002 - 625 - \frac{25 ( 250 2 - 125 )}{3}

x y^{^{'}} + (11 + x) y = e^{- x}

\int \frac{x ^{3}}{x ^{2} + 36} d x

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

\int arccos (4 x) d x

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

What is the integral from 0 to 5 of x^3sqrt(x^2+25) ?
The integral from 0 to 5 of x^3sqrt(x^2+25) is 2500sqrt(2)-625-(25(250sqrt(2)-125))/3