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

The integral from 0 to 2 of xsqrt(6-x^2) is (2(3sqrt(6)-sqrt(2)))/3

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

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

\frac{2 ( 3 6 - 2 )}{3}

Decimal Notation

3.95617 \dots

Solution steps

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

Apply u-substitution

: \int_{6}^{2} - \frac{u}{2} d u

= \int_{6}^{2} - \frac{u}{2} d u

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

= - \int_{2}^{6} - \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_{2}^{6} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{6}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{6})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{6}

Compute the boundaries:

46 - \frac{4 2}{3}

= \frac{1}{2} (46 - \frac{4 2}{3})

Simplify

= \frac{2 ( 3 6 - 2 )}{3}

\int \frac{x ^{2}}{4 - x ^{3}} d x

x = 1 \sum \infty \frac{x}{x ^{2} + 1}

\int_{0}^{7} w^{2} ln (w) d w

\int cot^{3} (t) d t

y = - 3 cos (x), at x = \frac{- π}{4}

What is the integral from 0 to 2 of xsqrt(6-x^2) ?
The integral from 0 to 2 of xsqrt(6-x^2) is (2(3sqrt(6)-sqrt(2)))/3