What is the integral from 0 to 4 of xsqrt(19-x^2) ?

The integral from 0 to 4 of xsqrt(19-x^2) is (19sqrt(19)-3sqrt(3))/3

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integral from 0 to 4 of xsqrt(19-x^2)

\int_{0}^{4} x 19 - x^{2} d x

\frac{19 19 - 3 3}{3}

Decimal Notation

25.87430 \dots

Solution steps

\int_{0}^{4} x 19 - x^{2} d x

Apply u-substitution

: \int_{19}^{3} - \frac{u}{2} d u

= \int_{19}^{3} - \frac{u}{2} d u

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

= - \int_{3}^{19} - \frac{u}{2} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{3}^{19} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{19}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{19})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{19}

Compute the boundaries:

\frac{38 19}{3} - 23

= \frac{1}{2} (\frac{38 19}{3} - 23)

Simplify

= \frac{19 19 - 3 3}{3}

\int_{0}^{\frac{1}{2}} \frac{4}{3} (1 - x^{3}) d x

\int_{0}^{10} x^{2} d x

\frac{d}{d x} (3 + e^{- 3 x} - 3 e^{- 3 x} x)

\frac{d}{d x} (\frac{- 11}{2 ( x - 2 ) ^{2}} - \frac{4}{x - 2})

\frac{2 s + 3}{s ^{2} + 17}

What is the integral from 0 to 4 of xsqrt(19-x^2) ?
The integral from 0 to 4 of xsqrt(19-x^2) is (19sqrt(19)-3sqrt(3))/3