Mathematics

# Let $\displaystyle \frac{d}{dx}F\left ( x \right )=\frac{e^{\sin x}}{x},x> 0.$ If $\displaystyle \int_{1}^{4}\frac{2e^{\sin x^{2}}}{x}dx=F\left ( k \right )-F\left ( 1 \right )$ then one of the possible values of $\displaystyle k$ is

$16$

##### SOLUTION
$\int _{ 1 }^{ 4 }{ \cfrac { 2{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =F\left( k \right) -F\left( 1 \right)$
Substitute ${ x }^{ 2 }=t\Rightarrow 2xdx=dt$
$\int _{ 1 }^{ 16 }{ 2\cfrac { { e }^{ \sin { t } } }{ t } \cfrac { dt }{ 2 } } =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow \int _{ 1 }^{ 16 }{ \cfrac { { e }^{ \sin { t } } }{ t } dt } =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow { \left[ F\left( t \right) \right] }_{ 1 }^{ 16 }=F\left( k \right) -F\left( 1 \right) \\ \Rightarrow F\left( 16 \right) -F\left( 1 \right) =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow k=16$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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