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

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

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

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

\frac{2 2 - 1}{3}

Decimal

0.60947 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{4 2}{3} - \frac{2}{3}

= \frac{1}{2} (\frac{4 2}{3} - \frac{2}{3})

Simplify

= \frac{2 2 - 1}{3}

\int_{0}^{1} z e^{- z} d z

\int_{0}^{1} x - 1 d x

\int_{0}^{a} (x a^{2} - x^{2}) d x

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

\int_{1}^{x} sin (t) d t

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