Question

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

• A. 15
• B. 16
• C. 63
• D. 64

Answer 1

The correct option is D 64

$\frac{d}{dx}F\left(x\right)=\left(\frac{e^{sinx}}{x}\right),x>0$
$F\left(x\right)=\int \frac{e^{sinx}}{x}dx--------\left(1\right)$
$Also,{\int }_1^4\frac{3}{x}{e}^{sin{x}^3}dx={\int }_1^4\frac{3x^2}{x^3}.e^{sin{x}^3}dx=F\left(k\right)-F\left(1\right)\left(given\right)$
$Let{x}^3=z\Rightarrow 3x^{2}dx=dz$
$\therefore {\int }_1^{64}\frac{e^{sinz}}{z}dz=F\left(k\right)-F\left(1\right)$
$[Fromeq.(1\left)\right]$
$\Rightarrow \left[F\right(z){]}_1^{64}=F(k)-F(1)\Rightarrow k=64$