What is the integral from 0 to 2 of xsqrt(5-x^2) ?

The integral from 0 to 2 of xsqrt(5-x^2) is (5sqrt(5)-1)/3

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integral from 0 to 2 of xsqrt(5-x^2)

\int_{0}^{2} x 5 - x^{2} d x

\frac{5 5 - 1}{3}

Decimal Notation

3.39344 \dots

Solution steps

\int_{0}^{2} x 5 - x^{2} d x

Apply u-substitution

: \int_{5}^{1} - \frac{u}{2} d u

= \int_{5}^{1} - \frac{u}{2} d u

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

= - \int_{1}^{5} - \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}^{5} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Compute the boundaries:

\frac{10 5}{3} - \frac{2}{3}

= \frac{1}{2} (\frac{10 5}{3} - \frac{2}{3})

Simplify

= \frac{5 5 - 1}{3}

\frac{( s + 2 )}{( s ^{2} + s + 1 )}

\frac{- 25}{s ^{2} + 6 s + 25}

f (x) = x^{2} - 8, at x = 4

x y^{^{'}} = 3 y - 6 x^{2}

\int cos (π x) cos (4 π x) d x

What is the integral from 0 to 2 of xsqrt(5-x^2) ?
The integral from 0 to 2 of xsqrt(5-x^2) is (5sqrt(5)-1)/3