What is the integral from 8 to 16 of xsqrt(16x-x^2) ?

The integral from 8 to 16 of xsqrt(16x-x^2) is 512/3+128pi

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integral from 8 to 16 of xsqrt(16x-x^2)

\int_{8}^{16} x 16 x - x^{2} d x

\frac{512}{3} + 128 π

Decimal Notation

572.79052 \dots

Solution steps

\int_{8}^{16} x 16 x - x^{2} d x

Complete the square

16 x - x^{2} : - (x - 8)^{2} + 64

= \int_{8}^{16} x - (x - 8)^{2} + 64 d x

Apply u-substitution

: \int_{0}^{8} (u + 8) - u^{2} + 64 d u

= \int_{0}^{8} (u + 8) - u^{2} + 64 d u

Expand

(u + 8) - u^{2} + 64 : u - u^{2} + 64 + 8 - u^{2} + 64

= \int_{0}^{8} u - u^{2} + 64 + 8 - u^{2} + 64 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}^{8} u - u^{2} + 64 d u + \int_{0}^{8} 8 - u^{2} + 64 d u

\int_{0}^{8} u - u^{2} + 64 d u = \frac{512}{3}

\int_{0}^{8} 8 - u^{2} + 64 d u = 128 π

= \frac{512}{3} + 128 π

\frac{( 5 x - 25 )}{x ^{2} - 5 x}

y^{^{''}} + 4 y = - 12 sec (2 t)

y = 4 - 2 x - x^{- 3}

x \to - 2 lim (\frac{x - 2}{x + 2})

y^{^{'}} = \frac{x e ^{x^{2}}}{2}

What is the integral from 8 to 16 of xsqrt(16x-x^2) ?
The integral from 8 to 16 of xsqrt(16x-x^2) is 512/3+128pi