What is the integral from 0 to 2 of 2xsqrt(1+3x^2) ?

The integral from 0 to 2 of 2xsqrt(1+3x^2) is (26sqrt(13)-2)/9

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integral from 0 to 2 of 2xsqrt(1+3x^2)

\int_{0}^{2} 2 x 1 + 3 x^{2} d x

\frac{26 13 - 2}{9}

Decimal

10.19381 \dots

Solution steps

\int_{0}^{2} 2 x 1 + 3 x^{2} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x 1 + 3 x^{2} d x

Apply u-substitution

: \int_{1}^{13} \frac{u}{6} d u

= 2 \cdot \int_{1}^{13} \frac{u}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int_{1}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

= 2 \cdot \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

Simplify

2 \cdot \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13} : \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{2}{3}

= \frac{1}{3} (\frac{26 13}{3} - \frac{2}{3})

Simplify

= \frac{26 13 - 2}{9}

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What is the integral from 0 to 2 of 2xsqrt(1+3x^2) ?
The integral from 0 to 2 of 2xsqrt(1+3x^2) is (26sqrt(13)-2)/9