Question

Suppose that F(x) is an anti-derivative of $f\left(x\right)=\frac{\sin x}{x}$, where $x>0$.

Then ${\int }_1^3\frac{\sin 2x}{x}dx$ 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

Answer: (A)

$F\left(6\right)-F\left(2\right)$

Let $I={\int }_1^3\frac{sin2x}{x}dx$

Substitute $2x=t\Rightarrow 2dx=dt$

Gives $I={\int }_2^6\frac{sint}{t}dt$

As antiderivative of $f\left(x\right)=\frac{sinx}{x}$ is $F\left(x\right)$

$I={\left[F\left(x\right)\right]}_2^6=F\left(6\right)-F\left(2\right)$