Find-0-5-1-2-x-2-2x-1-dx-

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} - 2 x + 1 = {(x - 1)}^{2} \dots

\sqrt{x^{2}} = {\begin{cases} x; x ⩾ 0 \\ - x ⩽ 0 \end{cases}

Answered by mathzup last updated on 21/Jun/24

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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