What is the integral from 0 to 1 of 20rsqrt(1-r^2) ?

The integral from 0 to 1 of 20rsqrt(1-r^2) is 20/3

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integral from 0 to 1 of 20rsqrt(1-r^2)

\int_{0}^{1} 20 r 1 - r^{2} d r

\frac{20}{3}

Decimal

6.66666 \dots

Solution steps

\int_{0}^{1} 20 r 1 - r^{2} d r

Take the constant out:

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

= 20 \cdot \int_{0}^{1} r 1 - r^{2} d r

Apply u-substitution

= 20 \cdot \int_{1}^{0} - u^{2} d u

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

= 20 (- \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

= 20 (- (- \int_{0}^{1} u^{2} d u))

Apply the Power Rule

= 20 (- (- [\frac{u ^{3}}{3}]_{0}^{1}))

Simplify

= 20 [\frac{u ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

= 20 \cdot \frac{1}{3}

Simplify

= \frac{20}{3}

What is the integral from 0 to 1 of 20rsqrt(1-r^2) ?
The integral from 0 to 1 of 20rsqrt(1-r^2) is 20/3