Question

Find the points of extrema of \[f(x) = \int\limits_0^x {\dfrac{{\sin t}}{t}dt} \] in the domain x>0.
A.\[n\pi ;n = 1,2,...\]
B. \[(2n + 1)\dfrac{\pi }{2};n = 1,2,...\]
C. \[(4n + 1)\dfrac{\pi }{2};n = 1,2,...\]
D. \[(2n + 1)\dfrac{\pi }{2};n = 1,2, < ..\]

Answer 1

Hints First differentiate the given function with respect to x, then differentiate the obtained function twice, equate the first derivative to zero and obtain the critical points. Substitute the critical points on the second derivative to observe the sign of the resultant.

Formula used
\[\dfrac{d}{{dx}}\int {f(x)} dx = f(x)\]
\[\dfrac{d}{{dx}}(\sin x) = \cos x\]
\[\dfrac{d}{{dx}}(\dfrac{1}{x}) = - \dfrac{1}{{{x^2}}}\]
\[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}\]
The general solution of \[\sin x = 0\] is \[x = n\pi ,n = 1,2,3,...\]

Complete step by step solution
Differentiate the given function \[f(x) = \int\limits_0^x {\dfrac{{\sin t}}{t}dt} \] with respect to x,
\[f'(x) = \dfrac{{\sin x}}{x}\]
Equate \[f'(x)\] to zero, to obtain the critical points.
\[\dfrac{{\sin x}}{x} = 0\]
\[\sin x = 0\]
\[x = n\pi ,n = 1,2,3,...\]
Now, differentiate the function \[f'(x) = \dfrac{{sinx}}{x}\] with respect to x.
\[f''(x) = \dfrac{{x.\dfrac{d}{{dx}}(sinx) - \sin x\dfrac{d}{{dx}}(x)}}{{{x^2}}}\]
  \[ = \dfrac{{x.\cos x - \sin x.1}}{{{x^2}}}\]
  \[ = \dfrac{{x\cos x - \sin x}}{{{x^2}}}\]
Substitute \[x = n\pi \] in the equation \[f''(x) = \dfrac{{x\cos x - \sin x}}{{{x^2}}}\] for evaluation.
\[f''(n\pi ) = \dfrac{{n\pi \cos n\pi - \sin n\pi }}{{{{(n\pi )}^2}}}\]
Now, value of \[\sin n\pi \] is 0.
But \[\cos n\pi > 0\] when n is even and \[\cos n\pi < 0\]when n is odd.
Therefore,
\[f''(n\pi ) > 0\], when n is even
\[f''(n\pi ) < 0\], when n is odd
Therefore, at \[x = n\pi \] the given function has extreme value.
The correct option is A.

Note Sometime students know how to derive the extreme value but get confused with the given function as it is in integration form, so for this remember that integration is anti-derivative, when we differentiate an integration then we get the function itself only.