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Evaluate $ \int^4_1 e^ \sqrt {x}dx $

solution:-

Here $1<x<4$

$1<\sqrt x<2$

$e<e^ \sqrt {x}<e^ 2$

$\int^4_1 $e dx$<\int^4_1 e^ \sqrt {x}dx<\int^4_1 e^ 2dx$

$3e <\int^4_1 e^ \sqrt {x}dx<3 e^ 2 $

But in this objective question

Options are

a)$e $

b)$e^2 $

c)$2e $

d)$2e^2 $

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Hint: Make the substitution $x=t^2$ to get the integral $2\int_1^2 te^t \mathrm{d}t$.

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  • $\begingroup$ thanks a lot . your method is better than mine $\endgroup$ – rst Jun 3 '13 at 11:04
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HINT: $$ 3 e < 2e^2 < 3e^2 $$ As suggested, the substitution $x=t^2$ yields $$ 2\int_1^2 te^t dt $$ Using integration by parts, you get $$ 2\int_1^2 te^t dt = 2\left([te^t]_1^2 - \int_1^2 e^t dt\right) = 2 \left( 2e^2 - e^1 -e^2 + e^1 \right) = 2 e^2 $$

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  • $\begingroup$ So you mean ans is (d)$2e^2$ $\endgroup$ – rst Jun 3 '13 at 10:51
  • $\begingroup$ Yes. Do you know how to calculate? $\endgroup$ – UrošSlovenija Jun 3 '13 at 10:52
  • $\begingroup$ I think $2e^2 > 3e$ so $3e^2>2e^2 > 3e$ $\endgroup$ – rst Jun 3 '13 at 10:55
  • $\begingroup$ That is the same as in my answer but reversed :) $\endgroup$ – UrošSlovenija Jun 3 '13 at 10:58
  • $\begingroup$ Thanks. But I have to choose only one ans $\endgroup$ – rst Jun 3 '13 at 11:03

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