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

The integral from 0 to 4 of xsqrt(4-x) is 128/15

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

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

\frac{128}{15}

Solution steps

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

Apply u-substitution

: \int_{4}^{0} - (- u + 4) u d u

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

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

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

Take the constant out:

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

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

Expand

(- u + 4) u : - u^{\frac{3}{2}} + 4 u

= - (- \int_{0}^{4} - u^{\frac{3}{2}} + 4 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}^{4} u^{\frac{3}{2}} d u + \int_{0}^{4} 4 u d u))

\int_{0}^{4} u^{\frac{3}{2}} d u = \frac{64}{5}

\int_{0}^{4} 4 u d u = \frac{64}{3}

= - (- (- \frac{64}{5} + \frac{64}{3}))

Simplify

- (- (- \frac{64}{5} + \frac{64}{3})) : \frac{128}{15}

= \frac{128}{15}

x \to \infty lim (\frac{( x + 1 ) ^{2} e ^{x^{3}}}{e ^{(x + 1)^{3}} x ^{2}})

x \to 16 - lim (\frac{8}{11 + 2 ^{\frac{1}{x - 16}}})

\int 2 (x^{2} + 2 x)^{16} (2 x + 2) d x

\frac{2 s}{( s ^{2} + 2 s + 6 )}

f (x) = 3^{x^{2} - 1}, at x = 2

What is the integral from 0 to 4 of xsqrt(4-x) ?
The integral from 0 to 4 of xsqrt(4-x) is 128/15