Question

If we have the values as \[f\left( 0 \right)=1,f\left( 2 \right)=3,f'\left( 2 \right)=5\] and \[f'\left( 0 \right)\] is finite, then \[\int\limits_{0}^{1}{x.f''\left( 2x \right)dx}\] is equal to: -
(a) 0
(b) 1
(c) 2
(d) None of these

Answer 1

Hint: Apply integration by parts rule by assuming x as function 1 and \[f''\left( 2x \right)\] as function 2. Use the formula: - \[\int\limits_{a}^{b}{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)dx}=\left[ {{f}_{1}}\left( x \right).\int{{{f}_{2}}\left( x \right)dx} \right]_{a}^{b}-\int\limits_{a}^{b}{\left( \int{{{f}_{2}}\left( x \right)dx} \right).f_{1}^{'}\left( x \right)}\], where ‘b’ and ‘a’ are the upper and lower limits of the given integral respectively and \[f_{1}^{'}\left( x \right)=\dfrac{d\left[ {{f}_{1}}\left( x \right) \right]}{dx}\]. Now, apply the same integration by parts rule one more time and simplify the expression by substituting the limits.

Complete step-by-step solution
Here, we have been provided with the integral: - \[\int\limits_{0}^{1}{x.f''\left( 2x \right)dx}\] and we have to find its value with some given information.
Now, here we can see that we have a product of two functions x and \[f''\left( 2x \right)\], so we have to apply integration by parts method. In this method we assume the two functions as function 1 (\[{{f}_{1}}\left( x \right)\]) and function 2 (\[{{f}_{2}}\left( x \right)\]) and apply the formula given as: - \[\Rightarrow \int\limits_{a}^{b}{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)dx}=\left[ {{f}_{1}}\left( x \right).\int{{{f}_{2}}\left( x \right)dx} \right]_{a}^{b}-\int\limits_{a}^{b}{\left( \int{{{f}_{2}}\left( x \right)dx} \right).f_{1}^{'}\left( x \right)}\], where ‘b’ and ‘a’ are lower and upper limits of the integral respectively and \[f_{1}^{'}\left( x \right)=\dfrac{d\left[ {{f}_{1}}\left( x \right) \right]}{dx}\].
Now, in the above question we have a product of x and \[f''\left( 2x \right)\] and we have been provided with the values of \[f'\left( 0 \right),f'\left( 2 \right),f\left( 0 \right)\] and \[f\left( 2 \right)\]. That means we have to integrate \[f''\left( 2x \right)\] so it must be assumed \[{{f}_{2}}\left( x \right)\]. Therefore, we have,
\[{{f}_{1}}\left( x \right)=x\] and \[{{f}_{2}}\left( x \right)=f''\left( 2x \right)\]
So, applying integration by parts rule, we have,
\[\Rightarrow \int\limits_{0}^{1}{x.f''\left( 2x \right)dx}=\left[ x.\int{f''\left( 2x \right)dx} \right]_{0}^{1}-\int{\left\{ \left( \int{f''\left( 2x \right)dx} \right).\dfrac{d\left[ x \right]}{dx} \right\}dx}\]
Here, we have \[f''\left( 2x \right)\] in which the coefficient of x is 2, so \[\int{f''\left( 2x \right)}=\dfrac{1}{2}\left[ f'\left( 2x \right) \right]\]. Therefore, we get,
\[\begin{align}
  & \Rightarrow \int\limits_{0}^{1}{x.f''\left( 2x \right)dx}=\left[ x\times \dfrac{1}{2}f'\left( 2x \right) \right]_{0}^{1}-\dfrac{1}{2}\int{\left( f'\left( 2x \right)\times 1 \right)dx} \\
 & \Rightarrow \int\limits_{0}^{1}{x.f''\left( 2x \right)dx}=\dfrac{1}{2}\times \left[ xf'\left( 2x \right) \right]_{0}^{1}-\dfrac{1}{2}\times \dfrac{1}{2}\left[ f\left( 2x \right) \right]_{0}^{1} \\
\end{align}\]
Substituting the limits we get,
\[\Rightarrow \int\limits_{0}^{1}{x.f''\left( 2x \right)dx}=\dfrac{1}{2}\left[ 1\times f'\left( 2 \right)-0\times f'\left( 0 \right) \right]-\dfrac{1}{4}\left[ f\left( 2 \right)-f\left( 0 \right) \right]\]
Substituting the given values of \[f\left( 0 \right),f\left( 2 \right)\] and \[f'\left( 2 \right)\], we get,
\[\begin{align}
  & \Rightarrow \int\limits_{0}^{1}{x.f''\left( 2x \right)dx}=\dfrac{1}{2}\left[ 5 \right]-\dfrac{1}{4}\times \left[ 3-1 \right] \\
 & \Rightarrow \int\limits_{0}^{1}{x.f''\left( 2x \right)dx}=2 \\
\end{align}\]
Hence, option (c) is the correct answer.

Note: One may note that here we cannot apply the ILATE rule for the integration by parts methods because we don’t know about the function \[f''\left( 2x \right)\]. So, we must check the given information in the question to get the answer. You must remember the integration by parts to solve the question because we don’t have any other method to find the integration of the product of two functions.