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

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

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

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

2 - 2

Decimal Notation

0.58578 \dots

Solution steps

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

Simplify

\frac{2}{2 - r ^{2}} : \frac{2}{2 - r ^{2}}

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

Take the constant out:

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

= 2 \cdot \int_{0}^{1} \frac{1}{2 - r ^{2}} r d r

Simplify

\frac{1}{2 - r ^{2}} r : \frac{r}{2 - r ^{2}}

= 2 \cdot \int_{0}^{1} \frac{r}{2 - r ^{2}} d r

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

22 - 2

= \frac{1}{2} (22 - 2)

Simplify

= 2 - 2

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