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

The integral from 0 to 2 of 2pixsqrt(1+4x^2) is (pi(17sqrt(17)-1))/6

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

\int_{0}^{2} 2 π x 1 + 4 x^{2} d x

\frac{π ( 17 17 - 1 )}{6}

Decimal Notation

36.17690 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{17} \frac{u}{8} d u

= 2 π \cdot \int_{1}^{17} \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}^{17} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

= 2 π \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Simplify

2 π \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17} : \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

= \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Compute the boundaries:

\frac{34 17}{3} - \frac{2}{3}

= \frac{π}{4} (\frac{34 17}{3} - \frac{2}{3})

Simplify

= \frac{π ( 17 17 - 1 )}{6}

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What is the integral from 0 to 2 of 2pixsqrt(1+4x^2) ?
The integral from 0 to 2 of 2pixsqrt(1+4x^2) is (pi(17sqrt(17)-1))/6