What is the integral from 0 to 2 of ysqrt(2+4y^2) ?

The integral from 0 to 2 of ysqrt(2+4y^2) is (13sqrt(2))/3

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integral from 0 to 2 of ysqrt(2+4y^2)

\int_{0}^{2} y 2 + 4 y^{2} d y

\frac{13 2}{3}

Decimal Notation

6.12825 \dots

Solution steps

\int_{0}^{2} y 2 + 4 y^{2} d y

Apply u-substitution

: \int_{2}^{18} \frac{u}{8} d u

= \int_{2}^{18} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{2}^{18} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{18}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{18}

Compute the boundaries:

362 - \frac{4 2}{3}

= \frac{1}{8} (362 - \frac{4 2}{3})

Simplify

= \frac{13 2}{3}

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What is the integral from 0 to 2 of ysqrt(2+4y^2) ?
The integral from 0 to 2 of ysqrt(2+4y^2) is (13sqrt(2))/3