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

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

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

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

\frac{3 3 - 1}{6}

Solution steps

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

Apply u-substitution

: \int_{3}^{1} - \frac{u}{4} d u

= \int_{3}^{1} - \frac{u}{4} d u

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

= - \int_{1}^{3} - \frac{u}{4} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{1}^{3} u d u)

Apply the Power Rule

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

= - (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{3})

Simplify

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{3}

Compute the boundaries:

23 - \frac{2}{3}

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

Simplify

= \frac{3 3 - 1}{6}

\int_{- 1}^{2} (x + 1) - (x^{2} - 1) d x

\int_{0}^{\frac{π}{2}} 3 sin (y) cos (y) d y

\int_{- 1}^{1} (2 - x^{2}) - x^{2} d x

\int_{0}^{1} 1 + 2 x d x

\int_{0}^{3} 25 - x^{2} d x

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