Question

Suppose that $F\left(x\right)$ is an anti - derivative of $f\left(x\right)=\frac{\sin x}{x}$, where $x>0,$ then $\underset{1}{\overset{3}{\int }}\frac{\sin 2x}{x}dx$ can be expressed as

• A. $F\left(6\right)-F\left(2\right)$
• B. $\frac{1}{2}\left(F\right(6)-F(2\left)\right)$
• C. $\frac{1}{2}\left(F\right(3)-F(1\left)\right)$
• D. $2\left(F\right(6)-F(2\left)\right)$

Answer 1

The correct option is A $F\left(6\right)-F\left(2\right)$

Let $I=\underset{1}{\overset{3}{\int }}\frac{\sin 2x}{x}dx$
Put $2x=t,\Rightarrow dx=\frac{dt}{2}$
$\Rightarrow I=\frac{2}{2}\underset{2}{\overset{6}{\int }}\frac{\sin t}{t}dt=\underset{2}{\overset{6}{\int }}\frac{\sin t}{t}dt$
But given $\int \frac{\sin x}{x}dx=F\left(x\right)$
$\Rightarrow \underset{2}{\overset{6}{\int }}\frac{\sin t}{t}dt=F\left(6\right)-F\left(2\right)$