What is the integral from 0 to 2 of ysqrt(x+2) ?

The integral from 0 to 2 of ysqrt(x+2) is ((16-4sqrt(2))y)/3

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integral from 0 to 2 of ysqrt(x+2)

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

\frac{( 16 - 4 2 ) y}{3}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{2}^{4} u d u

= y \cdot \int_{2}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{4}

= y [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{4}

Compute the boundaries:

\frac{16}{3} - \frac{4 2}{3}

= y (\frac{16}{3} - \frac{4 2}{3})

Simplify

= \frac{( 16 - 4 2 ) y}{3}

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What is the integral from 0 to 2 of ysqrt(x+2) ?
The integral from 0 to 2 of ysqrt(x+2) is ((16-4sqrt(2))y)/3