What is the integral from 0 to 19 of xsqrt(19-x) ?

The integral from 0 to 19 of xsqrt(19-x) is (1444sqrt(19))/(15)

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integral from 0 to 19 of xsqrt(19-x)

\int_{0}^{19} x 19 - x d x

\frac{1444 19}{15}

Decimal Notation

419.61667 \dots

Solution steps

\int_{0}^{19} x 19 - x d x

Apply u-substitution

: \int_{19}^{0} - (- u + 19) u d u

= \int_{19}^{0} - (- u + 19) u d u

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

= - \int_{0}^{19} - (- u + 19) u d u

Take the constant out:

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

= - (- \int_{0}^{19} (- u + 19) u d u)

Expand

(- u + 19) u : - u^{\frac{3}{2}} + 19 u

= - (- \int_{0}^{19} - u^{\frac{3}{2}} + 19 u d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{0}^{19} u^{\frac{3}{2}} d u + \int_{0}^{19} 19 u d u))

\int_{0}^{19} u^{\frac{3}{2}} d u = \frac{722 19}{5}

\int_{0}^{19} 19 u d u = \frac{722 19}{3}

= - (- (- \frac{722 19}{5} + \frac{722 19}{3}))

Simplify

- (- (- \frac{722 19}{5} + \frac{722 19}{3})) : \frac{1444 19}{15}

= \frac{1444 19}{15}

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What is the integral from 0 to 19 of xsqrt(19-x) ?
The integral from 0 to 19 of xsqrt(19-x) is (1444sqrt(19))/(15)