Let, ddxf(x)=esin⁡xx,x>0 .If ∫14 3xesin⁡x3dx=f(k)−f(1) then the possible value of k, is

ddxf(x)=esin⁡xx⇒∫14 3xesin⁡x3dx=∫14 3x2x3esin⁡x3dx Put x3=t⇒3x2dx=dtf(t)=∫164 esin⁡ttdt=∫164 f(t)dt=f(64)−f(1)On comparing, we getk=64

Let, ddxf(x)=esin⁡xx,x>0 .If ∫14 3xesin⁡x3dx=f(k)−f(1) then the possible value of k, is

ddxf(x)=esin⁡xx⇒∫14 3xesin⁡x3dx=∫14 3x2x3esin⁡x3dx Put x3=t⇒3x2dx=dtf(t)=∫164 esin⁡ttdt=∫164 f(t)dt=f(64)−f(1)On comparing, we getk=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 the possible value of k, is

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Detailed Solution

$\begin{array}{l} \frac{d}{dx} f (x) = \frac{e^{\sin x}}{x} \Rightarrow \int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} dx = \int_{1}^{4} \frac{3 x^{2}}{x^{3}} e^{\sin x^{3}} dx \\ Put x^{3} = t \Rightarrow 3 x^{2} dx = dt \\ f (t) = \int_{1}^{64} \frac{e^{\sin t}}{t} dt = \int_{1}^{64} f (t) dt = f (64) - f (1) \end{array}$