clculate ∫_0 ^1 x(√(x^2 −2x+2)) dx


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Question Number 31069 by abdo imad last updated on 02/Mar/18

$c l c u l a t e \int_{0}^{1} x \sqrt{x^{2} - 2 x + 2} d x$
Commented by abdo imad last updated on 03/Mar/18

$l e t p u t I = \int_{0}^{1} x \sqrt{x^{2} - 2 x + 2 d x}$ $I = \int_{0}^{1} x \sqrt{{(x - 1)}^{2} + 1} d x t h e c h . x - 1 = s h t$ $I = \int_{a r g s h (- 1)}^{o} (1 + s h t) c h t c h t d t = - \int_{0}^{l n (- 1 + \sqrt{2})} {c h}^{2} t d t$ $- \int_{0}^{l n (- 1 + \sqrt{2})} s h t {c h}^{2} t d t b u t w e h a v e$ $\int_{0}^{l n (- 1 + \sqrt{2})} {c h}^{2} t d t = \int_{0}^{l n (- 1 + \sqrt{2})} \frac{1 + c h (2 t)}{2} d t$ $= \frac{1}{2} l n (- 1 + \sqrt{2}) + \frac{1}{4} {[s h (2 t)]}_{0}^{l n (- 1 + \sqrt{2})}$ $= \frac{1}{2} l n (- 1 + \sqrt{2}) + \frac{1}{4} s h (2 l n (- 1 + \sqrt{2})) a n d$ $\int_{0}^{l n (- 1 + \sqrt{2})} s h t {c h}^{2} t d t = \frac{1}{3} {[{c h}^{3} t]}_{0}^{l n (- 1 + \sqrt{2})}$ $= \frac{1}{3} ({c h}^{3} (l n (- 1 + \sqrt{2}) - 1) b y u s i n g s h u = \frac{e^{u} - e^{- u}}{2} a n d$ $c h u = \frac{e^{u} + e^{- u}}{2} t b e v a l u e o f I i s d e t e r m i n e d .$
	Answered by Joel578 last updated on 02/Mar/18

	$I = \int_{0}^{1} (x - 1 + 1) \sqrt{x^{2} - 2 x + 2} d x$ $= \int_{0}^{1} (x - 1) \sqrt{x^{2} - 2 x + 2} d x + \int_{0}^{1} \sqrt{{(x - 1)}^{2} + 1} d x$ $I_{1} = \int (x - 1) \sqrt{x^{2} - 2 x + 2} d x$ $u = x^{2} - 2 x + 2 \to d u = 2 (x - 1) d x$ $I_{1} = \frac{1}{2} \int (x - 1) \sqrt{u} . \frac{d u}{x - 1}$ $= \frac{1}{3} u \sqrt{u} = \frac{1}{3} {(x^{2} - 2 x + 2)}^{\frac{3}{2}} + C$ $I_{2} = \int \sqrt{{(x - 1)}^{2} + 1} d x$ $x - 1 = \tan θ \to d x = \sec^{2} θ d θ$ $I_{2} = \int \sqrt{\tan^{2} θ + 1} . \sec^{2} θ d θ$ $= \int \sec^{3} θ d θ$ $= \frac{\tan θ \sec θ}{2} - \frac{1}{2} \ln ∣ \sec θ + \tan θ ∣ + C$ $= \frac{(x - 1) \sqrt{x^{2} - 2 x + 2}}{2} - \frac{1}{2} \ln ∣ \sqrt{x^{2} - 2 x + 2} + x - 1 ∣ + C$ $I = {[I_{1} + I_{2}]}_{0}^{1}$ $= {[\frac{1}{3} {(x^{2} - 2 x + 2)}^{\frac{3}{2}} + \frac{(x - 1) \sqrt{x^{2} - 2 x + 2}}{2} - \frac{1}{2} \ln ∣ \sqrt{x^{2} - 2 x + 2} + x - 1 ∣]}_{0}^{1}$ $= (\frac{1}{3} + 0 - 0) - (\frac{2 \sqrt{2}}{3} - \frac{\sqrt{2}}{2} - \frac{1}{2} \ln (\sqrt{2} - 1))$ $= \frac{1 - 2 \sqrt{2}}{3} + \frac{1}{2} (\sqrt{2} + \ln (\sqrt{2} - 1))$
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