What is the integral from 0 to 6 of 7xsqrt(20-x^2) ?

The integral from 0 to 6 of 7xsqrt(20-x^2) is (280sqrt(5))/3+i 448/3

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integral from 0 to 6 of 7xsqrt(20-x^2)

\int_{0}^{6} 7 x 20 - x^{2} d x

\frac{280 5}{3} + i \frac{448}{3}

Solution steps

\int_{0}^{6} 7 x 20 - x^{2} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{6} x 20 - x^{2} d x

Apply u-substitution

: \int_{20}^{- 16} - \frac{u}{2} d u

= 7 \cdot \int_{20}^{- 16} - \frac{u}{2} d u

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

= 7 (- \int_{- 16}^{20} - \frac{u}{2} d u)

Take the constant out:

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

= 7 (- (- \frac{1}{2} \cdot \int_{- 16}^{20} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{- 16}^{20}

= 7 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 16}^{20}))

Simplify

7 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 16}^{20})) : \frac{7}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 16}^{20}

= \frac{7}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 16}^{20}

Compute the boundaries:

\frac{80 5}{3} + i \frac{128}{3}

= \frac{7}{2} (\frac{80 5}{3} + i \frac{128}{3})

Simplify

= \frac{280 5}{3} + i \frac{448}{3}

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What is the integral from 0 to 6 of 7xsqrt(20-x^2) ?
The integral from 0 to 6 of 7xsqrt(20-x^2) is (280sqrt(5))/3+i 448/3