Question

If \[{\text{f(0) = 1,f(2) = 3,f'(2) = 5}}\] and \[{\text{f'(0)}}\] is finite, then \[\int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} \] is equal to ?
A. Zero
B. \[1\]
C. \[2\]
D. \[{\text{3}}\]

Answer 1

Hint: Firstly progressing the question with the approach of the application of integration by parts and then putting up the limits in it and if it will be necessary to apply by parts again then do so and proceed by putting the values and obtain the desired answer.

Complete step by step answer:

As per the given values are \[{\text{f(0) = 1,f(2) = 3,f'(2) = 5}}\] and \[{\text{f'(0)}}\] is finite, and we need to find the solution of \[\int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} \]
\[\int {uvdx = u\int {vdx - \int {\dfrac{{du}}{{dx}}} } } (\int {vdx} )dx\] is the formula for by parts.
Let \[u = x\] and \[v = f''(x)\],
So, applying by parts in \[\int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} \], we get,
\[\int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} {\text{ = x}}\int\limits_0^1 {f''(2x)} {\text{dx - }}\int\limits_{\text{0}}^{\text{1}} {\dfrac{{dx}}{{dx}}(\int\limits_0^1 {f''(2x} ){\text{dx)dx}}} \]
On simplifying we get,
\[\int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} {\text{ = (}}\dfrac{{\text{1}}}{{\text{2}}}{\text{)xf'(2x)|}}_{\text{0}}^{\text{1}}{\text{ - }}\int\limits_{\text{0}}^{\text{1}} {\dfrac{{{\text{f'(2x)}}}}{{\text{2}}}{\text{.dx}}} \]
On applying limits we get,
\[ \Rightarrow \int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} {\text{ = (}}\dfrac{{\text{1}}}{{\text{2}}}{\text{)(f'(2) - 0) - }}\int\limits_{\text{0}}^{\text{1}} {\dfrac{{{\text{f'(2x)}}}}{{\text{2}}}{\text{.dx}}} \]
So, after applying by parts one time we can simplify it up to
\[ \Rightarrow \int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} {\text{ = (}}\dfrac{{\text{1}}}{{\text{2}}}{\text{)f'(2) - }}\int\limits_{\text{0}}^{\text{1}} {\dfrac{{{\text{f'(2x)}}}}{{\text{2}}}{\text{.dx}}} \]
Now integrating the second part of the above equation, we get,
\[\dfrac{1}{2}\int\limits_0^1 {f'(2x).dx} = \dfrac{1}{2}(\dfrac{{(f(2x))}}{2})|_0^1\]
Now, substitute this in the above equation again, and on substituting the limits we get,
\[\int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} = (\dfrac{1}{2})f'(2) - \dfrac{1}{4}(f(2) - f(0))\]
Now, substituting \[{\text{f(0) = 1,f(2) = 3,f'(2) = 5}}\] in the equation, we get,
\[
  \int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} = (\dfrac{5}{2}) - \dfrac{1}{4}(3 - 1) \\
   = (\dfrac{5}{2}) - \dfrac{1}{2} \\
   = \dfrac{4}{2} \\
   = 2 \\
  \]
Hence, \[\int\limits_{\text{0}}^{\text{1}} {{\text{xf''(2x)dx}}} = 2\].
Hence, option (C) is the correct answer.

Note: In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative.
Properly apply the By- parts rule and if possible substitute and calculate as all the values of the final equation are known.
By parts rule is \[\int {uvdx = u\int {vdx - \int {\dfrac{{du}}{{dx}}} } } (\int {vdx} )dx\]