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Let $\frac{\mathrm{d} }{\mathrm{d} x}f(x)$ = $\frac{e^{sinx}}{x}, x>0$ if $\int_{1}^{4}\frac{2e^{sinx^{2}}}{x}d(x)$ = f(k)-f(1) then k = ______
asked in Calculus by Boss (7.2k points)  
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f(k) - f(1)

= $\int_{-\infty }^{k}\frac{e^{sinx}}{x}$ - $\int_{-\infty}^{1}\frac{e^{sinx}}{x}$

= $\int_{1}^{k}\frac{e^{sinx}}{x}$    ...................(1)

 

Now, we simplify the integral

I = $\int_{1}^{4}\frac{2e^{sinx^{2}}}{x}$

 

Put x2 = t

$\therefore 2x dx=dt$

$\therefore dx=\frac{dt}{2\sqrt{t}}$

 

$\therefore$ I = $\int_{1}^{16} \frac{2e^{sint}dt}{\sqrt{t}*2\sqrt{t}}$   ...........(2)

 

Now, when we simplify statement (2), we get statement (1).

 

$\therefore k=16$

answered by Veteran (14.8k points)  
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Two errors:

1. Statement 2 should be integration from 1 to 16.

2. Statement 2 should be integration of $\frac{2e^{sint}}{\sqrt{t}*2\sqrt{t}}$
Thanks . will edit

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