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

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

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

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{0} - \frac{u}{2} d u

= 2 \cdot \int_{4}^{0} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{0}^{4} - \frac{u}{2} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{2} \cdot \int_{0}^{4} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= 2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}))

Simplify

2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})) : [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

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