Question

Let $\frac{d}{dx}F\left(x\right)=\frac{e^{\sin x}}{x},x>0$. If ${\int }_1^4\frac{2e^{\sin x^2}}{x}dx=F\left(k\right)-F\left(1\right)$ then one of the possible value of k is

Answer 1

${\int }_1^4\frac{2e^{\sin x^2}}{x}dx=F\left(k\right)-F\left(1\right)$

Put $x^2=t\Rightarrow 2xdx=dt$

${\int }_1^{16}2\frac{e^{\sin t}}{t}\frac{dt}{2}=F\left(k\right)-F\left(1\right)$

$\Rightarrow {\int }_1^{16}\frac{e^{\sin t}}{t}dt=F\left(k\right)-F\left(1\right)$

$\Rightarrow {\left[F\left(t\right)\right]}_1^{16}=F\left(k\right)-F\left(1\right)$

$\Rightarrow F\left(16\right)-F\left(1\right)=F\left(k\right)-F\left(1\right)$

$\Rightarrow k=16$