What is the integral from 0 to 5 of 3xsqrt(25-x^2) ?

The integral from 0 to 5 of 3xsqrt(25-x^2) is 125

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integral from 0 to 5 of 3xsqrt(25-x^2)

\int_{0}^{5} 3 x 25 - x^{2} d x

125

Solution steps

\int_{0}^{5} 3 x 25 - x^{2} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{5} x 25 - x^{2} d x

Apply u-substitution

: \int_{25}^{0} - \frac{u}{2} d u

= 3 \cdot \int_{25}^{0} - \frac{u}{2} d u

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

= 3 (- \int_{0}^{25} - \frac{u}{2} d u)

Take the constant out:

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

= 3 (- (- \frac{1}{2} \cdot \int_{0}^{25} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25}

= 3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25}))

Simplify

3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25})) : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25}

= \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25}

Compute the boundaries:

\frac{250}{3}

= \frac{3}{2} \cdot \frac{250}{3}

Simplify

= 125

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What is the integral from 0 to 5 of 3xsqrt(25-x^2) ?
The integral from 0 to 5 of 3xsqrt(25-x^2) is 125