What is the integral from 0 to 1 of xsqrt(6-x^2) ?

The integral from 0 to 1 of xsqrt(6-x^2) is (6sqrt(6)-5sqrt(5))/3

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integral from 0 to 1 of xsqrt(6-x^2)

\int_{0}^{1} x 6 - x^{2} d x

\frac{6 6 - 5 5}{3}

Solution steps

\int_{0}^{1} x 6 - x^{2} d x

Apply u-substitution

: \int_{6}^{5} - \frac{u}{2} d u

= \int_{6}^{5} - \frac{u}{2} d u

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

= - \int_{5}^{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_{5}^{6} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{6}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{6})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{6}

Compute the boundaries:

46 - \frac{10 5}{3}

= \frac{1}{2} (46 - \frac{10 5}{3})

Simplify

= \frac{6 6 - 5 5}{3}

\frac{d}{d x} (sin (x) + \frac{3}{4} cot (x))

\frac{\partial}{\partial x} (cos (x y^{2}))

y^{^{''}} + 4 y = e^{(- t)}

csc (x + 5)

\frac{d y}{d x} - (\frac{1 + x}{x}) y = \frac{y ^{2}}{x}

What is the integral from 0 to 1 of xsqrt(6-x^2) ?
The integral from 0 to 1 of xsqrt(6-x^2) is (6sqrt(6)-5sqrt(5))/3