∫_( 0) ^( 2) ((x^2 +3x)/(√(x+2))) dx = ?


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Question Number 84055 by naka3546 last updated on 09/Mar/20

$\int_{0} \overset{2}{} \frac{x^{2} + 3 x}{\sqrt{x + 2}} d x = ?$
Commented by mathmax by abdo last updated on 09/Mar/20
$A =∫_0 ^2 ((x^2 +3x)/(√(x+2)))dx we do the changement (√(x+2))=t ⇒x+2=t^2 A =∫_(√2) ^2 (((t^2 −2)^2 +3(t^2 −2))/t)(2t)dt =2 ∫_(√2) ^2 { t^4 −4t^2 +4 +3t^2 −6)dt =2[(t^5 /5)−(4/3)t^3 +t^3 −2t]_(√2) ^2 =2{ ((32)/5)−((32)/3) +4 −((((√2))^5 )/5)+(4/3)((√2))^3 −2(√2)+2(√2)}=...$
$A = \int_{0}^{2} \frac{x^{2} + 3 x}{\sqrt{x + 2}} d x w e d o t h e c h a n g e m e n t \sqrt{x + 2} = t \Rightarrow x + 2 = t^{2}$ $A = \int_{\sqrt{2}}^{2} \frac{{(t^{2} - 2)}^{2} + 3 (t^{2} - 2)}{t} (2 t) d t$ $= 2 \int_{\sqrt{2}}^{2} {t^{4} - 4 t^{2} + 4 + 3 t^{2} - 6) d t$ $= 2 {[\frac{t^{5}}{5} - \frac{4}{3} t^{3} + t^{3} - 2 t]}_{\sqrt{2}}^{2}$ $= 2 {\frac{32}{5} - \frac{32}{3} + 4 - \frac{{(\sqrt{2})}^{5}}{5} + \frac{4}{3} {(\sqrt{2})}^{3} - 2 \sqrt{2} + 2 \sqrt{2}} = . . .$
	Answered by john santu last updated on 09/Mar/20

	$let \sqrt{x + 2} = t \Rightarrow x = t^{2} - 2$ $\int \underset{\sqrt{2}}{\overset{2}{}} \frac{(t^{2} - 2) (t^{2} + 1) 2 t dt}{t}$ $= 2 \int \underset{\sqrt{2}}{\overset{2}{}} (t^{4} - t^{2} - 2) dt$ $= 2 {[\frac{1}{5} t^{5} - \frac{1}{3} t^{3} - 2 t]}_{\sqrt{2}}^{2}$ $= 2 [\frac{1}{5} (32 - 4 \sqrt{2}) - \frac{1}{3} (8 - 2 \sqrt{2}) - 2 (2 - \sqrt{2})]$
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