Question

Let $\frac{d}{dx}F\left(x\right)=\frac{e^{smx}}{x}$ , $x>0$. lf ${\int }_1^4\frac{3}{x}{e}^{sm{x}^3}dx=F\left(k\right)-F\left(1\right)$, then one of the possible values of $k$ is

• A. $16$
• B. $62$
• C. $64$
• D. $15$

Answer 1

The correct option is C $64$

Given: ${\int }_1^4\frac{3}{x}{e}^{\sin x^3}dx$

Let $z=x^3⟹dz=3x^{2}dx=\frac{3}{x}{x}^{3}dx$

$⟹dz=\frac{3}{x}zdx$

$⟹\frac{3}{x}dx=\frac{dz}{z}$

And when $x=1,z=1$ and when $x=4,z=64$

Hence, integration becomes:-

${\int }_1^{64}{e}^{\sin z}\frac{dz}{z}$

Now it is given that:- $\frac{d}{dx}F\left(x\right)=\frac{e^{\sin x}}{x},x>0$

$⟹{\int }_1^{64}\frac{e^{\sin z}}{z}dz$

$={\int }_1^{64}dF\left(z\right)$

$=F\left(64\right)-F\left(1\right)$

Hence, $k=64$