Question

Let y=f(x) be a quadratic function with f '(2)=1. Find the value of the integral $ \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} $

Answer 1

Hint: In this question, we need to evaluate the value of the definite integral $ \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} $ . For this, we will use the property of the quadratic equation and the follow the integral by parts method.

Complete step-by-step answer:
The standard quadratic equation is given as: $ f(x) = a{x^2} + bx + c $
Differentiating the standard quadratic equation, we get: $ f'(x) = ax + b $
Again, differentiating the equation, we get: $ f''(x) = a $
So, we can say that the double differentiation of the quadratic equation yields a constant value.
Now, following the integral by parts to integrate the integral $ \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} $ where, f(x) is considered to be the first function and $ \sin \left( {\dfrac{{x - 2}}{2}} \right) $ is considered to be the second function.
 $
  \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} = f(x) \times \int {\sin \left( {\dfrac{{x - 2}}{2}} \right)dx} - \int {\left( {\dfrac{d}{{dx}}\left( {f(x)} \right) \times \int {\sin \left( {\dfrac{{x - 2}}{2}} \right)dx} } \right)dx} \\
   = f(x) \times \dfrac{{ - \cos \left( {\dfrac{{x - 2}}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} - \int {f'(x) \times } \left( {\dfrac{{ - \cos \left( {\dfrac{{x - 2}}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}} \right)dx \\
   = - 2f(x)\cos \left( {\dfrac{{x - 2}}{2}} \right) + 2\left[ {\int {f'(x)\cos \left( {\dfrac{{x - 2}}{2}} \right)dx} } \right] \\
  $
Again, applying the integral by parts method on the above equation by considering f’(x) as the first function and $ \cos \left( {\dfrac{{x - 2}}{2}} \right) $ as the second function, we get
 $
\int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} = - 2f(x)\cos \left( {\dfrac{{x - 2}}{2}} \right) + 2\left[ {f'(x)\int {\cos \left( {\dfrac{{x - 2}}{2}} \right)dx - \int {\left( {\dfrac{d}{{dx}}\left( {f'(x)} \right) \times \int {\cos \left( {\dfrac{{x - 2}}{2}} \right)dx} } \right)dx} } } \right] \\
= - 2f(x)\cos \left( {\dfrac{{x - 2}}{2}} \right) + 2f'(x) \times \dfrac{{\sin \left( {\dfrac{{x - 2}}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} - 2\int {f''(x) \times \dfrac{{\sin \left( {\dfrac{{x - 2}}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}} dx \\
  $
As, the double differentiation of the quadratic equation, is given by $ f''(x) = a $ so, we can take the constant term out of the integral part.
 $
  \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} = - 2f(x)\cos \left( {\dfrac{{x - 2}}{2}} \right) + 4f'(x)\sin \left( {\dfrac{{x - 2}}{2}} \right) + 4f''(x)\dfrac{{\cos \left( {\dfrac{{x - 2}}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} \\
   = - 2f(x)\cos \left( {\dfrac{{x - 2}}{2}} \right) + 4f'(x)\sin \left( {\dfrac{{x - 2}}{2}} \right) + 8f''(x)\cos \left( {\dfrac{{x - 2}}{2}} \right) \\
  $
According to the question, f’(2)=1 and lower and upper limits are $ \left( {2 - \pi } \right) $ and $ \left( {2 + \pi } \right) $ so, substituting the value in the above equation, we get
\[\int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} = \left| { - 2f(x)\cos \left( {\dfrac{{x - 2}}{2}} \right) + 4f'(x)\sin \left( {\dfrac{{x - 2}}{2}} \right) + 8f''(x)\cos \left( {\dfrac{{x - 2}}{2}} \right)} \right|_{2 - \pi }^{2 + \pi }\]
The upper limit is calculated as:
\[
  UL = \left( { - 2f(2 + \pi )\cos \left( {\dfrac{{2 + \pi - 2}}{2}} \right) + 4f'(2 + \pi )\sin \left( {\dfrac{{2 + \pi - 2}}{2}} \right) + 8f''(2 + \pi )\cos \left( {\dfrac{{2 + \pi - 2}}{2}} \right)} \right) \\
   = \left( { - 2f(2 + \pi )\cos \dfrac{\pi }{2} + 4f'(2 + \pi )\sin \dfrac{\pi }{2} + 8f''(2 + \pi )\cos \dfrac{\pi }{2}} \right) \\
   = 4f'(2 + \pi ) \\
 \]

Similarly, the lower limit is:
\[
  LL = \left( { - 2f(2 - \pi )\cos \left( {\dfrac{{2 - \pi - 2}}{2}} \right) + 4f'(2 - \pi )\sin \left( {\dfrac{{2 - \pi - 2}}{2}} \right) + 8f''(2 - \pi )\cos \left( {\dfrac{{2 - \pi - 2}}{2}} \right)} \right) \\
   = \left( { - 2f(2 - \pi )\cos \dfrac{\pi }{2} + 4f'(2 - \pi )\sin \left( {\dfrac{{ - \pi }}{2}} \right) + 8f''(2 - \pi )\cos \left( {\dfrac{{ - \pi }}{2}} \right)} \right) \\
   = - 4f'(2 - \pi ) \\
 \]
Hence, we can write
 $
  \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} = UL - LL \\
   = 4f'(2 + \pi ) + 4f'(2 - \pi ) \\
  $

Substituting the value of f’(x)=ax+b in the above equation, we get
\[
  \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} = 4f'(2 + \pi ) + 4f'(2 - \pi ) \\
   = 4\left[ {a(2 + \pi ) + b + \left( {a(2 - \pi ) + b} \right)} \right] \\
   = 4\left( {2a + a\pi + b + 2a - a\pi + b} \right) \\
   = 4\left( {4a + 2b} \right) \\
   = 8(2a + b) \\
 \]
Substituting the value of the x as 2 in the equation $ f'(x) = ax + b $ , we get
 $
  f'(x) = ax + b \\
  f'(2) = 2a + b \\
   = 1 \\
  $
So, the above equation can be written as:
\[
  \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} = 8 \times 1 \\
   = 8 \\
 \]
Hence, the value of the definite integral $ \int\limits_{2 - \pi }^{2 + \pi } {f(x) \cdot \sin \left( {\dfrac{{x - 2}}{2}} \right)dx} $ is 8.

Note: Students should be careful while considering the values of the first and the second function by following the ILATE rule which is expanded as Inverse, Logarithmic, Algebraic, Trigonometric and Exponential. The function which comes first in ILATE should be taken as the first function.