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

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

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

\int_{0}^{2} (x 4 - x^{2}) d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{0} - \frac{u}{2} d u

= \int_{4}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{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_{0}^{4} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

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

Simplify

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

Compute the boundaries:

\frac{16}{3}

= \frac{1}{2} \cdot \frac{16}{3}

Simplify

= \frac{8}{3}

\frac{d}{d x} (2 + 3 x)

- 2 x^{2} + 8, 3 x + 1, 0

\int (x^{2} - 7 x + 4) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = (5 + x) e^{- 2 x}

\frac{d y}{d x} + \frac{y}{x} = e^{x}

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