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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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$\int \limits_1^4 \large\frac{2 e^{\Large \sin x^2}}{x}$$=f(k)-f(1)$ then find $k$=? if $\large\frac{d}{dx}$$f(x)=\Large\frac{e^{\sin x }}{\normalsize x}$

\[\begin {array} {1 1} (a)\;k=14 \\ (b)\;k=16 \\ (c)\;k=10 \\ (d)\;k=18 \end {array}\]

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1 Answer

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$\int \limits_1^4 \large\frac{2 e^{\Large \sin x^2}}{x}.$$dx$
=> $x^2 =t=>x=1,t=1$
=>$2xdx=dt=> x=4,t=16$
$\int \limits_1^{16} \large\frac{2.e^{\Large \sin t}}{2.x^2}.$$dt$
=>$\int \limits_1^{16} \large\frac{e^{\Large \sin t}}{t}.$$dt$
=>$\int \limits_1^{16} \large\frac{e^{\Large \sin x}}{x}.$$dx$
=>$\int \limits_1^{16} \large\frac{e^{\Large \sin x}}{x}$
we given : $\large\frac{d(f(x))}{dx}=\frac{e^{\sin x}}{x}$
=> $ f(x) =\int \large\frac{e^{\sin x}}{x}$
=> $ [f(x) ]_1^{16}$
=> $[f(16)-f(1)]$
=> $[f(k)-f(1)]$
$= [f(16)-f(1)]$
$k=16$
Hence b is the correct answer.
answered Dec 12, 2013 by meena.p
 

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