What is the integral from 0 to 1 of xsqrt(3-x^2) ?

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

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integral from 0 to 1 of xsqrt(3-x^2)

\int_{0}^{1} x 3 - x^{2} d x

\frac{3 3 - 2 2}{3}

Decimal Notation

0.78924 \dots

Solution steps

\int_{0}^{1} x 3 - x^{2} d x

Apply u-substitution

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

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

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

= - \int_{2}^{3} - \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}^{3} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

23 - \frac{4 2}{3}

= \frac{1}{2} (23 - \frac{4 2}{3})

Simplify

= \frac{3 3 - 2 2}{3}

f (x) = [1 + e^{x}] [ln (x)], at x = 1

f (x) = e^{x} cos (x)

v^{''} - v = 0

\int t e^{- s t} sin (t) d t

\frac{\partial}{\partial x} (- \frac{x cos ( y )}{x - 1})

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