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

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

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

d er i v a t i v e y = 4 - 2 x - x^{- 3}

\int r^{2} sec^{2} (r^{3}) d r

t an g e n t - 6 + 250 x^{3} - 3 x^{5}

\frac{d}{d x} (\frac{1}{5 x + 1})

x \to - 1 f lim (x)

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