Question

If $f\left( x \right) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt} $ then which of the following is true?
A) $f'''\left( x \right) + f'\left( x \right) = \cos x - 2x\sin x$
B) $f'''\left( x \right) + f'\left( x \right) - f'\left( x \right) = \cos x$
C) $f'''\left( x \right) - f''\left( x \right) = \cos x - 2x\sin x$
D) $f'''\left( x \right) + f''\left( x \right) = \sin x$

Answer 1

Hint:Observe the function carefully and express it as two different integrals. Don’t try to solve any of the integrals instead use the fundamental theorem of calculus to solve the problem. Note the expression for each of the derivatives carefully.

Complete step-by-step answer:
Observe the function carefully. It is the difference of the two functions.
We know that integration of the difference of two functions is the same as the difference of the integrals of the respective functions provided both the functions are integrable.
We will open up the bracket in the given function and express the given function as follow:
$f\left( x \right) = \int\limits_0^x {\left( {t\sin x - t\sin t} \right)dt} $
Now we will separate the integrals as follows:
$f\left( x \right) = \int\limits_0^x {t\sin xdt} - \int\limits_0^x {t\sin tdt} $
Observe that the first integral is with respect to $t$ therefore, we will treat $\sin x$ as constant while evaluating the integral.
Therefore, the function can be rewritten as:
$f\left( x \right) = \sin x\int\limits_0^x {tdt} - \int\limits_0^x {t\sin tdt} $
Now the fundamental theorem of calculus states that if $f$ is a continuous real valued differentiable function defined on the closed interval $\left[ {a,b} \right]$ such that we define another function $F\left( x \right)$ as $F\left( x \right) = \int\limits_a^x {f\left( t \right)} dt$ then $F'\left( x \right) = f\left( x \right)$ for all $x \in \left( {a,b} \right)$ .
Using the fundamental we will differentiate the function $f(x)$ as follows:
$f'\left( x \right) = \sin x\left( x \right) + \cos x\int\limits_0^x {tdt} - x\sin x$ …….… (1)
Note that to evaluate the first derivative we also used the product rule of differentiation.
The product rule states that for two differentiable functions $u,v$ the derivative of the product is written as follows:
$\left( {uv} \right)' = u'v + v'u$
Observe that the equation (1) can be further simplified as follows:
$f'\left( x \right) = \cos x\int\limits_0^x {tdt} $ ………… (2)
Differentiate the equation (2) with respect to $x$ again as follows:
$f''\left( x \right) = x\cos x - \sin x\int\limits_0^x {tdt} $ … (3)
Note that we again used fundamental theorem as well as the product rule while differentiating the above function.
If you observe the options then you see that the options contain the derivatives till order three. So, we will differentiate the equation (3) again with respect to $x$ as follows:
$f'''\left( x \right) = - x\sin x + \cos x - x\sin x - \cos x\int\limits_0^x {tdt} $
This can be further simplified as follows:
$f'''\left( x \right) = \cos x - 2x\sin x - \cos x\int\limits_0^x {tdt} $ … (4)
Now observe that from the equation (2) we can write $f'\left( x \right) = \cos x\int\limits_0^x {tdt} $ .
Therefore, the equation (4) can be further rewritten as:
$f'''\left( x \right) = \cos x - 2x\sin x - f'\left( x \right)$
Rearrange the terms and write the following:
$f'''\left( x \right) + f'\left( x \right) = \cos x - 2x\sin x$

So, the correct answer is “Option A”.

Note:We have used the fundamental theorem of calculus to make the problem simpler. One can actually solve the integral but that will create unnecessary complications. Also check why the other options are not correct.