What is the integral from 0 to 1 of sqrt(4-r^2)r ?

The integral from 0 to 1 of sqrt(4-r^2)r is (8-3sqrt(3))/3

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integral from 0 to 1 of sqrt(4-r^2)r

\int_{0}^{1} 4 - r^{2} r d r

\frac{8 - 3 3}{3}

Solution steps

\int_{0}^{1} 4 - r^{2} r d r

Apply u-substitution

: \int_{4}^{3} - \frac{u}{2} d u

= \int_{4}^{3} - \frac{u}{2} d u

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

= - \int_{3}^{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_{3}^{4} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{16}{3} - 23

= \frac{1}{2} (\frac{16}{3} - 23)

Simplify

= \frac{8 - 3 3}{3}

\int_{0}^{1} (14 ln (x)) d x

\int_{- 1}^{2} x^{2} - 4 d x

\int_{- 1}^{1} - z^{2} + 1 d z

\int_{- π}^{π} sin (θ) d θ

\int_{- π}^{π} π d x

What is the integral from 0 to 1 of sqrt(4-r^2)r ?
The integral from 0 to 1 of sqrt(4-r^2)r is (8-3sqrt(3))/3