Suppose F(x) is an antiderivative of f(x)=sinxx, x>0 then ∫13sin2xxdx can be expressed as

∫sinxxdx=Fx+cI=∫13sin2xxdx Put 2x=t ⇒ dx=dt2 ⇒ I=∫26sinttdt=F(6)−F(2)

Suppose F(x) is an antiderivative of f(x)=sinxx, x>0 then ∫13sin2xxdx can be expressed as

∫sinxxdx=Fx+cI=∫13sin2xxdx Put 2x=t ⇒ dx=dt2 ⇒ I=∫26sinttdt=F(6)−F(2)

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Suppose $F (x)$ is an antiderivative of $f (x) = \frac{\sin x}{x}, x > 0$ then $\int_{1}^{3} \frac{\sin 2 x}{x} d x$ can be expressed as

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$\frac{1}{2} (F (6) - F (2))$

$(F (6) - F (2))$

$\frac{1}{2} (F (3) - F (1))$

$2 (F (6) - F (2))$

answer is B.

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Detailed Solution

$\int \frac{\sin x}{x} d x = F (x) + c$

$I = \int_{1}^{3} \frac{\sin 2 x}{x} d x P u t 2 x = t \Rightarrow d x = \frac{d t}{2} \Rightarrow I = \int_{2}^{6} \frac{\sin t}{t} d t = F (6) - F (2)$

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Suppose F(x) is an antiderivative of f(x)=sinxx, x>0 then ∫13sin2xxdx can be expressed as

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Suppose $F (x)$ is an antiderivative of $f (x) = \frac{\sin x}{x}, x > 0$ then $\int_{1}^{3} \frac{\sin 2 x}{x} d x$ can be expressed as