Let frac{d}{dx}F(x) = frac{e^{sin x}}{x} for | vedclass

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(D) Given that $\frac{d}{dx}F(x) = \frac{e^{\sin x}}{x}$.We need to evaluate the integral $I = \int_{1}^{4} \frac{3}{x} e^{\sin(x^3)} dx$.Multiply the

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$64$

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(D) Given that $\frac{d}{dx}F(x) = \frac{e^{\sin x}}{x}$.We need to evaluate the integral $I = \int_{1}^{4} \frac{3}{x} e^{\sin(x^3)} dx$.Multiply the numerator and denominator by $x^2$ to facilitate substitution:$I = \int_{1}^{4} \frac{3x^2}{x^3} e^{\sin(x^3)} dx$.Let $t = x^3$,then $dt = 3x^2 dx$.When $x = 1$,$t = 1^3 = 1$.When $x = 4$,$t = 4^3 = 64$.Substituting these into the integral:$I = \int_{1}^{64} \frac{e^{\sin t}}{t} dt$.Since $\frac{d}{dt}F(t) = \frac{e^{\sin t}}{t}$,the integral becomes:$I = [F(t)]_{1}^{64} = F(64) - F(1)$.Comparing this with the given expression $F(k) - F(1)$,we get $k = 64$.

Let $\frac{d}{dx}F(x) = \frac{e^{\sin x}}{x}$ for $x > 0$. 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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