Let ddxF(x)=esin⁡xx,x>0 if ∫14 3xesin⁡x3dx=F(k)−F(1) then one of the possible values of k is

ddx(F(x))=esinxx⇒F(x)=∫esin⁡xxdxF(k)−F(1)=∫14 3xesin⁡x3dx = ∫14 1x3esin⁡x3dx3 put x3=t ,if x=1⇒t=1 and x=4⇒t=43=64 =∫164 esin⁡ttdt=[F(t)]164=F(64)−F(1)∴K=64

Let ddxF(x)=esin⁡xx,x>0 if ∫14 3xesin⁡x3dx=F(k)−F(1) then one of the possible values of k is

ddx(F(x))=esinxx⇒F(x)=∫esin⁡xxdxF(k)−F(1)=∫14 3xesin⁡x3dx = ∫14 1x3esin⁡x3dx3 put x3=t ,if x=1⇒t=1 and x=4⇒t=43=64 =∫164 esin⁡ttdt=[F(t)]164=F(64)−F(1)∴K=64

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Let $\frac{d}{dx} F (x) = \frac{e^{\sin x}}{x}, x > 0 if \int_{1}^{4} \frac{3}{x} e^{\sin (x^{3})} dx = F (k) - F (1)$ then one of the possible values of k is

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Detailed Solution

$\begin{array}{l} \frac{d}{dx} (F (x)) = \frac{e^{sinx}}{x} \Rightarrow F (x) = \int \frac{e^{\sin x}}{x} dx \\ F (k) - F (1) = \int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} dx \\ = \int_{1}^{4} \frac{1}{x^{3}} e^{\sin x^{3}} d (x^{3}) \\ [put x^{3} = t, i f x = 1 \Rightarrow t = 1 and x = 4 \Rightarrow t = 4^{3} = 64] \\ = \int_{1}^{64} \frac{e^{\sin t}}{t} dt = [F (t)]_{1}^{64} \\ = F (64) - F (1) ∴ K = 64 \end{array}$