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

The integral from 0 to 2 of ysqrt(4-y^2) is 8/3

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

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

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

Apply u-substitution

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

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

= - \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

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

Apply the Power Rule

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

Simplify

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

Compute the boundaries:

\frac{16}{3}

= \frac{1}{2} \cdot \frac{16}{3}

Simplify

= \frac{8}{3}

\int_{0}^{4} (7 - x) x d x

\int_{0}^{1} x^{2} 4 x^{2} + 1 d x

\int_{1}^{2} \frac{y + 4 y ^{6}}{y ^{3}} d y

\int_{1}^{2 y} 2 x y d x

\int_{0}^{1} \frac{x + 2}{x ^{2} - x - 2} d x

What is the integral from 0 to 2 of ysqrt(4-y^2) ?
The integral from 0 to 2 of ysqrt(4-y^2) is 8/3