What is the integral from 0 to 1 of 2pixsqrt(1+4x^2) ?

The integral from 0 to 1 of 2pixsqrt(1+4x^2) is (pi(5sqrt(5)-1))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 2pixsqrt(1+4x^2)

\int_{0}^{1} 2 π x 1 + 4 x^{2} d x

\frac{π ( 5 5 - 1 )}{6}

Decimal

5.33041 \dots

Solution steps

\int_{0}^{1} 2 π x 1 + 4 x^{2} d x

Take the constant out:

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

= 2 π \cdot \int_{0}^{1} x 1 + 4 x^{2} d x

Apply u-substitution

: \int_{1}^{5} \frac{u}{8} d u

= 2 π \cdot \int_{1}^{5} \frac{u}{8} d u

Take the constant out:

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

= 2 π \frac{1}{8} \cdot \int_{1}^{5} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= 2 π \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Simplify

2 π \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5} : \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Compute the boundaries:

\frac{10 5}{3} - \frac{2}{3}

= \frac{π}{4} (\frac{10 5}{3} - \frac{2}{3})

Simplify

= \frac{π ( 5 5 - 1 )}{6}

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

\int_{0}^{8} \frac{3 x}{1 + x} d x

\int_{0}^{2 π} 3 sin (x) d x

\int_{1}^{5} ∣ 4 x - 8 ∣ d x

\int_{x}^{6} sin (t^{4}) d t

What is the integral from 0 to 1 of 2pixsqrt(1+4x^2) ?
The integral from 0 to 1 of 2pixsqrt(1+4x^2) is (pi(5sqrt(5)-1))/6