What is the integral from 0 to 2 of xsqrt(8-2x^2) ?

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

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integral from 0 to 2 of xsqrt(8-2x^2)

\int_{0}^{2} x 8 - 2 x^{2} d x

\frac{8 2}{3}

Decimal Notation

3.77123 \dots

Solution steps

\int_{0}^{2} x 8 - 2 x^{2} d x

Apply u-substitution

: \int_{8}^{0} - \frac{u}{4} d u

= \int_{8}^{0} - \frac{u}{4} d u

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

= - \int_{0}^{8} - \frac{u}{4} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{0}^{8} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{8}

= - (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{8})

Simplify

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{8}

Compute the boundaries:

\frac{32 2}{3}

= \frac{1}{4} \cdot \frac{32 2}{3}

Simplify

= \frac{8 2}{3}

\int \frac{2}{35 sin ( x ) - 12 cos ( x )} d x

\int x arctan (x^{2}) d x

\frac{2 t ^{2} + 1}{t}

\frac{d}{d x} (x e^{- x})

f (x) = cos (4 x)

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