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

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

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

\int_{0}^{1} (x 2 - x^{2}) d x

\frac{2 2 - 1}{3}

Decimal Notation

0.60947 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{1} - \frac{u}{2} d u

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

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

= - \int_{1}^{2} - \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_{1}^{2} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{4 2}{3} - \frac{2}{3}

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

Simplify

= \frac{2 2 - 1}{3}

y = \frac{x}{( 1 + x ^{2} ) , at ( 3 , 0.3 )}

\frac{d}{d x} (1 + 4 sin (x))

\int - 2 y - y^{2} + 2 d y

f (x) = 1 - 7 x^{2}, (- 2, - 27)

\frac{\partial}{\partial x} (e^{x y} + \frac{x}{y})

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