Suppose that F(x) is an antiderivative of f(x | vedclass

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(A) Given that $F(x)$ is an antiderivative of $f(x) = \frac{\sin x}{x}$,we have $\int \frac{\sin x}{x} dx = F(x) + C$. We need to evaluate the integra

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$F(6) - F(2)$

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(A) Given that $F(x)$ is an antiderivative of $f(x) = \frac{\sin x}{x}$,we have $\int \frac{\sin x}{x} dx = F(x) + C$. We need to evaluate the integral $I = \int_{1}^{3} \frac{\sin 2x}{x} dx$. Let $t = 2x$,then $dt = 2 dx$,which implies $dx = \frac{dt}{2}$. When $x = 1$,$t = 2$. When $x = 3$,$t = 6$. Substituting these into the integral: $I = \int_{2}^{6} \frac{\sin t}{t/2} \cdot \frac{dt}{2} = \int_{2}^{6} \frac{\sin t}{t} dt$. Since $\int \frac{\sin t}{t} dt = F(t)$,by the Fundamental Theorem of Calculus,$\int_{2}^{6} \frac{\sin t}{t} dt = F(6) - F(2)$. Thus,the correct option is $A$.

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

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