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

The integral from 0 to 1 of r^2sqrt(1-r^2) is pi/(16)

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

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

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (4 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

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

= \frac{1}{8} (\frac{π}{2} - 0)

Simplify

\frac{1}{8} (\frac{π}{2} - 0) : \frac{π}{16}

= \frac{π}{16}

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\int_{1}^{4} \frac{2 + 5 x}{x} d x

What is the integral from 0 to 1 of r^2sqrt(1-r^2) ?
The integral from 0 to 1 of r^2sqrt(1-r^2) is pi/(16)