Find: ∫_0 ^( 5) (√((1/2) (x^2 − 2x + 1))) dx = ?


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Question Number 208670 by hardmath last updated on 20/Jun/24

$Find : \int_{0}^{5} \sqrt{\frac{1}{2} (x^{2} - 2 x + 1)} dx = ?$
Commented by essaad last updated on 20/Jun/24

	Answered by Berbere last updated on 20/Jun/24
	$x^2 −2x+1=(x−1)^2 ... (√x^2 )= { ((x;x≥0)),((−x≤0)) :}$
	$x^{2} - 2 x + 1 = {(x - 1)}^{2} . . .$ $\sqrt{x^{2}} = {\begin{cases} x; x ⩾ 0 \\ - x ⩽ 0 \end{cases}$
	Answered by mathzup last updated on 21/Jun/24
	$I=(1/( (√2)))∫_0 ^5 (√((x−1)^2 ))dx=(1/( (√2)))∫_0 ^5 ∣x−1∣dx =(1/( (√2)))∫_0 ^1 (1−x)dx +(1/( (√2)))∫_1 ^5 (x−1)dx =(1/( (√2)))[x−(x^2 /2)]_0 ^1 +(1/( (√2)))[(x^2 /2)−x]_1 ^5 =(1/( (√2)))×(1/2)+(1/( (√2))){((25)/2)−5−((1/2)−1)} =(1/(2(√2)))+(1/( (√2)))(((15)/2)+(1/2))=(1/(2(√2)))+(8/( (√2))) =(1/( (√2)))(8+(1/2))=((17)/(2(√2)))$
	$I = \frac{1}{\sqrt{2}} \int_{0}^{5} \sqrt{{(x - 1)}^{2}} d x = \frac{1}{\sqrt{2}} \int_{0}^{5} ∣ x - 1 ∣ d x$ $= \frac{1}{\sqrt{2}} \int_{0}^{1} (1 - x) d x + \frac{1}{\sqrt{2}} \int_{1}^{5} (x - 1) d x$ $= \frac{1}{\sqrt{2}} {[x - \frac{x^{2}}{2}]}_{0}^{1} + \frac{1}{\sqrt{2}} {[\frac{x^{2}}{2} - x]}_{1}^{5}$ $= \frac{1}{\sqrt{2}} \times \frac{1}{2} + \frac{1}{\sqrt{2}} {\frac{25}{2} - 5 - (\frac{1}{2} - 1)}$ $= \frac{1}{2 \sqrt{2}} + \frac{1}{\sqrt{2}} (\frac{15}{2} + \frac{1}{2}) = \frac{1}{2 \sqrt{2}} + \frac{8}{\sqrt{2}}$ $= \frac{1}{\sqrt{2}} (8 + \frac{1}{2}) = \frac{17}{2 \sqrt{2}}$
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