What is the integral from 0 to 16 of xsqrt(16-x) ?

The integral from 0 to 16 of xsqrt(16-x) is 4096/15

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integral from 0 to 16 of xsqrt(16-x)

\int_{0}^{16} x 16 - x d x

\frac{4096}{15}

Decimal Notation

273.06666 \dots

Solution steps

\int_{0}^{16} x 16 - x d x

Apply u-substitution

: \int_{16}^{0} - (- u + 16) u d u

= \int_{16}^{0} - (- u + 16) u d u

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

= - \int_{0}^{16} - (- u + 16) u d u

Take the constant out:

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

= - (- \int_{0}^{16} (- u + 16) u d u)

Expand

(- u + 16) u : - u^{\frac{3}{2}} + 16 u

= - (- \int_{0}^{16} - u^{\frac{3}{2}} + 16 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}^{16} u^{\frac{3}{2}} d u + \int_{0}^{16} 16 u d u))

\int_{0}^{16} u^{\frac{3}{2}} d u = \frac{2048}{5}

\int_{0}^{16} 16 u d u = \frac{2048}{3}

= - (- (- \frac{2048}{5} + \frac{2048}{3}))

Simplify

- (- (- \frac{2048}{5} + \frac{2048}{3})) : \frac{4096}{15}

= \frac{4096}{15}

\int_{0}^{\infty} \frac{x}{x ^{3} + 1} d x

x \to 3 lim (\frac{4 x - 5}{5 x - 1})

\int + x^{2} x^{3} + 27 d x

\frac{2 s e ^{- 3 s}}{3 s ^{2} + 27}

\frac{d}{d x} (\frac{1}{1 - x ^{2} - y})

What is the integral from 0 to 16 of xsqrt(16-x) ?
The integral from 0 to 16 of xsqrt(16-x) is 4096/15