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

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

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

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

\frac{8 - 3 3}{3}

Decimal Notation

0.93461 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{16}{3} - 23

= \frac{1}{2} (\frac{16}{3} - 23)

Simplify

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

\int_{1}^{9} (\frac{1}{2 x}) d x

\int_{0}^{3} 3 1 + 36 x^{2} d x

\int_{0}^{2} 16 - 8 x^{2} + x^{4} d x

\int_{\frac{π}{6}}^{\frac{π}{3}} tan^{3} (x) d x

\int_{0}^{\infty} \frac{1}{θ} d θ

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