Question

Suppose that F(x) is an antiderivative of f(x)$=\frac{\sin x}{x},x>0$ then ${\int }_1^3\frac{\sin 2x}{x}$ can be expressed as

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

Answer 1

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

$F\left(x\right)=\int \frac{\sin x}{x}dx$ Now $I={\int }_1^3\frac{\sin 2x}{x}dx[put2x=t]={\int }_2^6\frac{2}{2}\frac{\sin }{t}dt={\left[F\left(x\right)\right]}_2^6=F\left(6\right)-F\left(2\right)$