Let ddx(F(x))=esin⁡xx,x>0 if ∫14 2esin⁡x2xdx=F(k)−F(1) ), then one of the possible values of k is,

A
4
B
8
C
16
D
32

Solution:

We have,

$\frac{d}{d x} (F (x)) = \frac{e^{\sin x}}{x} \Rightarrow \int \frac{e^{\sin x}}{x} d x = F (x)$ …(i)

Now,

$\int_{1}^{4} \frac{e^{\sin x^{2}}}{x} d x = \int_{1}^{4} \frac{e^{\sin x^{2}}}{x^{2}} \cdot d (x^{2}) = \int_{1}^{16} \frac{e^{\sin t}}{t} d t$ , where $t = x^{2}$

$\int_{1}^{4} 2 \frac{e^{\sin x^{2}}}{x} d x = [F (t)]^{16} = F (16) - F (1)$ [using (i)]

hence , $k = 16$

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