What is the integral from 0 to 1 of xsqrt(1-1/4 x^2) ?

The integral from 0 to 1 of xsqrt(1-1/4 x^2) is (8-3sqrt(3))/6

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integral from 0 to 1 of xsqrt(1-1/4 x^2)

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

\frac{8 - 3 3}{6}

Solution steps

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

Simplify

x 1 - \frac{1}{4} x^{2} : \frac{1}{2} x 4 - x^{2}

= \int_{0}^{1} \frac{1}{2} x 4 - x^{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} x 4 - x^{2} d x

Apply u-substitution

: \int_{4}^{3} - \frac{u}{2} d u

= \frac{1}{2} \cdot \int_{4}^{3} - \frac{u}{2} d u

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

= \frac{1}{2} (- \int_{3}^{4} - \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} (- (- \frac{1}{2} \cdot \int_{3}^{4} u d u))

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{16}{3} - 23

= \frac{1}{4} (\frac{16}{3} - 23)

Simplify

= \frac{8 - 3 3}{6}

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What is the integral from 0 to 1 of xsqrt(1-1/4 x^2) ?
The integral from 0 to 1 of xsqrt(1-1/4 x^2) is (8-3sqrt(3))/6