What is the integral from 0 to 1 of 2ysqrt(26+36y^2) ?

The integral from 0 to 1 of 2ysqrt(26+36y^2) is (31sqrt(62)-13sqrt(26))/(27)

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integral from 0 to 1 of 2ysqrt(26+36y^2)

\int_{0}^{1} 2 y 26 + 36 y^{2} d y

\frac{31 62 - 13 26}{27}

Decimal

6.58544 \dots

Solution steps

\int_{0}^{1} 2 y 26 + 36 y^{2} d y

Take the constant out:

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

= 2 \cdot \int_{0}^{1} y 26 + 36 y^{2} d y

Apply u-substitution

: \int_{26}^{62} \frac{u}{72} d u

= 2 \cdot \int_{26}^{62} \frac{u}{72} d u

Take the constant out:

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

= 2 \cdot \frac{1}{72} \cdot \int_{26}^{62} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{26}^{62}

= 2 \cdot \frac{1}{72} [\frac{2}{3} u^{\frac{3}{2}}]_{26}^{62}

Simplify

2 \cdot \frac{1}{72} [\frac{2}{3} u^{\frac{3}{2}}]_{26}^{62} : \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{26}^{62}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{26}^{62}

Compute the boundaries:

\frac{124 62 - 52 26}{3}

= \frac{1}{36} \cdot \frac{124 62 - 52 26}{3}

Simplify

= \frac{31 62 - 13 26}{27}

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What is the integral from 0 to 1 of 2ysqrt(26+36y^2) ?
The integral from 0 to 1 of 2ysqrt(26+36y^2) is (31sqrt(62)-13sqrt(26))/(27)