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Question: Suppose that F(x) is an antiderivative of \[f\left( x \right) = \dfrac{{\sin x}}{x},x > 0\] then \[\...

Suppose that F(x) is an antiderivative of f(x)=sin⁡xx,x>0f\left( x \right) = \dfrac{{\sin x}}{x},x > 0 then ∫13sin⁡2xx\int_1^3 {\dfrac{{\sin 2x}}{x}} can be expressed as
A) F(6)−F(2)F(6)-F(2)
B) 12(F(6)−F(2))\dfrac{1}{2}\left( {F\left( 6 \right) - F\left( 2 \right)} \right)
C) 12(F(3)−F(1))\dfrac{1}{2}\left( {F\left( 3 \right) - F\left( 1 \right)} \right)
D) 2(F(6)−F(2))2\left( {F\left( 6 \right) - F\left( 2 \right)} \right)

Explanation

Solution

Here we have been given antiderivative of f(x) which means nothing but integral. We can convert the question into the form of ∫sin⁡xx\int {\dfrac{{\sin x}}{x}} to get the solution in terms of F(x). We can carry out substitution to convert sin⁡2x\sin 2x to sin⁡x\sin x by using the intermediate variable as t so that we can simplify the integral to the desired form. After we get the integral in the desired form we will again resubstitute to get the integral in variable x.

Complete step-by-step solution
Let us start by simplifying the given information.
Antiderivative means nothing but integral of f(x).
So we can write F(x) is an antiderivative of f(x), mathematically as,
F(x)=∫f(x)dx=∫sin⁡xxdxF\left( x \right) = \int {f\left( x \right)dx = \int {\dfrac{{\sin x}}{x}dx} }
Now let us come to the integral which we need to simplify.
We have the integral as ∫13sin⁡2xx\int_1^3 {\dfrac{{\sin 2x}}{x}}
Let I=∫13sin⁡2xx\int_1^3 {\dfrac{{\sin 2x}}{x}}
Let us go for substitution by substituting 2x=t.
Let 2x=t, x=t2\dfrac{t}{2}
dx =dt2\dfrac{{dt}}{2}
The limits will also change as follows:
Lower limit: when x=1, t=2
Upper limit: when x=3, t=6
Therefore, now in terms of t, we have the integral as
I=∫262sin⁡ttdt2\int_2^6 {\dfrac{{2\sin t}}{t}\dfrac{{dt}}{2}}
We can simplify this further as,
I =∫26sin⁡ttdt\int_2^6 {\dfrac{{\sin t}}{t}dt}
We need to express the integral in the form of F(x), therefore we will convert above integral in terms of x.
I=∫26sin⁡xxdx\int_2^6 {\dfrac{{\sin x}}{x}dx}
Since,f(x)=sin⁡xxf\left( x \right) = \dfrac{{\sin x}}{x}, we can write,
I=∫26f(x)dx\int_2^6 {f\left( x \right)dx}
Now we already know that, F(x)=∫f(x)dx=∫sin⁡xxdxF\left( x \right) = \int {f\left( x \right)dx = \int {\dfrac{{\sin x}}{x}dx} } .
I=[F(x)]26\left[ {F\left( x \right)} \right]_2^6
Applying lower and upper limits, we get
=F(6)−F(2)=F(6)-F(2)
Therefore, the correct answer is option B.

Note: Remember to substitute 2x by t in order to get the equation in terms of sin⁡tt\dfrac{{\sin t}}{t} so that we can later convert it tosin⁡xx\dfrac{{\sin x}}{x}. Also, we have to keep in mind to change the limits when substituting x by t according to the relation between the two variables. For example, in this case, we are substituting 2x by t, hence the limits will consequently change from 1 and 3 to 2 and 6.