What is the integral from 0 to 2 of r^3sqrt(1+4r^2) ?

The integral from 0 to 2 of r^3sqrt(1+4r^2) is (391sqrt(17)+1)/(120)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of r^3sqrt(1+4r^2)

\int_{0}^{2} r^{3} 1 + 4 r^{2} d r

\frac{391 17 + 1}{120}

Decimal

13.44278 \dots

Solution steps

\int_{0}^{2} r^{3} 1 + 4 r^{2} d r

Apply u-substitution

: \int_{1}^{17} \frac{u ( u - 1 )}{32} d u

= \int_{1}^{17} \frac{u ( u - 1 )}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int_{1}^{17} u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{32} \cdot \int_{1}^{17} u^{\frac{3}{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

= \frac{1}{32} (\int_{1}^{17} u^{\frac{3}{2}} d u - \int_{1}^{17} u d u)

\int_{1}^{17} u^{\frac{3}{2}} d u = \frac{578 17 - 2}{5}

\int_{1}^{17} u d u = \frac{34 17 - 2}{3}

= \frac{1}{32} (\frac{578 17 - 2}{5} - \frac{34 17 - 2}{3})

Simplify

\frac{1}{32} (\frac{578 17 - 2}{5} - \frac{34 17 - 2}{3}) : \frac{391 17 + 1}{120}

= \frac{391 17 + 1}{120}

\int_{- 1}^{1} \frac{1}{e ^{5 t}} d t

\int_{1}^{3} 7 x^{2} d x

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

\int_{2}^{5} (x^{3} + 2) d x

\int_{0}^{0.5} \frac{1}{1 + x ^{4}} d x

What is the integral from 0 to 2 of r^3sqrt(1+4r^2) ?
The integral from 0 to 2 of r^3sqrt(1+4r^2) is (391sqrt(17)+1)/(120)