∫_( 1) ^4 e^(√x) dx =


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Question Number 83145 by 09658867628 last updated on 28/Feb/20

$\underset{1}{\int^{4}} e^{\sqrt{x}} d x =$
Commented by mathmax by abdo last updated on 28/Feb/20
$I=∫_1 ^4 e^(√x) dx changement (√x)=t give I=∫_1 ^2 e^t (2t)dt =2 ∫_1 ^2 t e^t dt =_(byparts) 2{ [te^t ]_1 ^2 −∫_1 ^2 e^t dt} =2{2e^2 −e −(e^2 −e)} =2{e^2 } ⇒I =2e^2$
$I = \int_{1}^{4} e^{\sqrt{x}} d x c h a n g e m e n t \sqrt{x} = t g i v e$ $I = \int_{1}^{2} e^{t} (2 t) d t = 2 \int_{1}^{2} t e^{t} d t =_{b y p a r t s} 2 {{[{t e}^{t}]}_{1}^{2} - \int_{1}^{2} e^{t} d t}$ $= 2 {2 e^{2} - e - (e^{2} - e)} = 2 {e^{2}} \Rightarrow I = 2 e^{2}$
	Answered by Kunal12588 last updated on 28/Feb/20
	$(√x)=t dx=2tdt I=2∫_1 ^2 te^t dt=2{[te^t ]_1 ^2 −∫_1 ^2 e^t dt} ⇒I=2{2e^2 −e−[e^t ]_1 ^2 } ⇒I=2{2e^2 −e−e^2 +e} ⇒I=2e^2 ∫_( 1) ^4 e^(√x) dx =2e^2 ; i hope this is correct$
	$\sqrt{x} = t$ $d x = 2 t d t$ $I = 2 \int_{1}^{2} {t e}^{t} d t = 2 {{[{t e}^{t}]}_{1}^{2} - \int_{1}^{2} e^{t} d t}$ $\Rightarrow I = 2 {2 e^{2} - e - {[e^{t}]}_{1}^{2}}$ $\Rightarrow I = 2 {2 e^{2} - e - e^{2} + e}$ $\Rightarrow I = 2 e^{2}$ $\underset{1}{\int^{4}} e^{\sqrt{x}} d x = 2 e^{2}; i h o p e t h i s i s c o r r e c t$
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