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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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}

\int \frac{ln ( x + 5 )}{x + 5} d x

\frac{\partial}{\partial t} (2)

\frac{d y}{d x} + y^{2} = 0

x \to 3 lim (\frac{2 ^{x} - 8}{x - 3})

\int \frac{8}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

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