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

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

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

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - u^{2} d u

= \int_{1}^{0} - u^{2} d u

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

= - \int_{0}^{1} - u^{2} d u

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{1}

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

Simplify

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

Compute the boundaries:

\frac{1}{3}

= \frac{1}{3}

\int (\frac{x ^{3}}{x - 1}) d x

\frac{d}{d x} ((1 - x)^{3} - 3 x (1 - x)^{2})

\frac{d y}{d x} = \frac{x - 1}{3 x ^{2} - 6 x - 1}

\int (ln (x)) 2 d x

\frac{ln ( x )}{ln ( 2 )}

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