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

The integral from 0 to 2 of xsqrt(2-x) is (16sqrt(2))/(15)

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

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

\frac{16 2}{15}

Decimal Notation

1.50849 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{0} - (- u + 2) u d u

= \int_{2}^{0} - (- u + 2) u d u

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

= - \int_{0}^{2} - (- u + 2) u d u

Take the constant out:

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

= - (- \int_{0}^{2} (- u + 2) u d u)

Expand

(- u + 2) u : - u^{\frac{3}{2}} + 2 u

= - (- \int_{0}^{2} - u^{\frac{3}{2}} + 2 u d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{0}^{2} u^{\frac{3}{2}} d u + \int_{0}^{2} 2 u d u))

\int_{0}^{2} u^{\frac{3}{2}} d u = \frac{8 2}{5}

\int_{0}^{2} 2 u d u = \frac{8 2}{3}

= - (- (- \frac{8 2}{5} + \frac{8 2}{3}))

Simplify

- (- (- \frac{8 2}{5} + \frac{8 2}{3})) : \frac{16 2}{15}

= \frac{16 2}{15}

\frac{d}{d x} ((8 x + 5) (x^{2} - 3 x))

\frac{d}{d x} (\frac{e ^{x}}{1 + x ^{2}})

\int_{1}^{6} \frac{2}{x ^{3}} d x

6 - 4 x^{2}

\frac{d y}{d x} - \frac{2}{3} x y = \frac{1}{3} y^{3}

What is the integral from 0 to 2 of xsqrt(2-x) ?
The integral from 0 to 2 of xsqrt(2-x) is (16sqrt(2))/(15)