∫_1 ^5 x^2 (√(x−1))dx=?


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 195515 by sulaymonnorboyev140 last updated on 04/Aug/23

$\underset{1}{\int^{5}} x^{2} \sqrt{x - 1} d x = ?$
	Answered by tri26112004 last updated on 04/Aug/23

	$S a y t = \sqrt{x - 1} \Rightarrow t^{2} = x - 1 \Leftrightarrow x^{2} = {(t^{2} + 1)}^{2}$ $\to d x = 2 t d t$ $∙ x = 1 t o 5 \Leftrightarrow t = 0 t o 2$ $= {\int_{0}}^{2} {(t^{2} + 1)}^{2} t d t$ $= {\int_{0}}^{2} (t^{5} + 2 t^{3} + t) d t$ $= (\frac{1}{6} t^{6} + \frac{1}{2} t^{4} + \frac{1}{2} t^{2}) ∣_{0}^{2} = . . .$
	Answered by Calculusboy last updated on 04/Aug/23

	$l e t u^{2} = x - 1 => u^{2} + 1 = x (i n s e r t t h e l i m i t)$ $2 u \frac{d u}{d x} = 1$ $d x = 2 u d u (c o n t i n u e d s i r)$
	Answered by Frix last updated on 04/Aug/23

	$\int x^{2} \sqrt{x - 1} d x \overset{[t = \sqrt{x - 1}]}{=} 2 \int t^{2} {(t^{2} + 1)}^{2} d t =$ $= \frac{2 t^{3} (15 t^{4} + 42 t^{2} + 35)}{105} =$ $= \frac{2 {(x - 1)}^{\frac{3}{2}} (15 x^{2} + 12 x + 8)}{105} + C$ $\int x^{2} \sqrt{x - 1} d x \overset{[t = \frac{1}{\sqrt{x - 1}}]}{=} - 2 \int \frac{{(t^{2} + 1)}^{2}}{t^{8}} d t =$ $= \frac{2 (35 t^{4} + 42 t^{2} + 15)}{105 t^{7}} =$ $= \frac{2 {(x - 1)}^{\frac{3}{2}} (15 x^{2} + 12 x + 8)}{105} + C$ $Answer is \frac{7088}{105}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum