Question

If $\int {{e^{2x}}f'\left( x \right)} dx = g\left( x \right)$, then the value of $\int {\left( {{e^{2x}}f'\left( x \right) + {e^{2x}}f\left( x \right)} \right)} dx$ is equal to:
(A) $\dfrac{1}{2}\left( {{e^{2x}}f\left( x \right) - g\left( x \right)} \right) + c$
(B) $\dfrac{1}{2}\left( {{e^{2x}}f\left( x \right) + g\left( x \right)} \right) + c$
(C) $\dfrac{1}{2}\left( {{e^{2x}}f\left( {2x} \right) + g\left( x \right)} \right) + c$
(D) $\dfrac{1}{2}\left( {{e^{2x}}f'\left( x \right) - g\left( x \right)} \right) + c$

Answer 1

Hint: In the given question, we are given the value of a specific integral as a function and we have to evaluate another integral in terms of the function. So, we will use the integration by parts method to simplify the integral and then substitute the value of the already given integral in it.

Complete answer:
So, we are given that $\int {{e^{2x}}f'\left( x \right)} dx = g\left( x \right)$.
We are required to find the value of $\int {\left( {{e^{2x}}f'\left( x \right) + {e^{2x}}f\left( x \right)} \right)} dx$.
We apply the method of integration by parts to simplify the integral. We first separate the two parts of the integration and then apply integration by parts in the first integral term. So, we get,
\[\int {{e^{2x}}f\left( x \right)dx + \int {{e^{2x}}f'\left( x \right)} } dx\]
In integration by parts method, we integrate a function which is a product of two functions using a formula:
$\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } $
So, applying this to the first integral term in equation $\left( 1 \right)$, we get,
\[ \Rightarrow f\left( x \right)\int {{e^{2x}}dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {{e^{2x}}dx} } \right]dx} } + \int {{e^{2x}}f'\left( x \right)} dx\]
Simplifying the expression, we get,
\[ \Rightarrow f\left( x \right)\int {{e^{2x}}dx - \int {\left[ {f'\left( x \right)\int {{e^{2x}}dx} } \right]dx} } + \int {{e^{2x}}f'\left( x \right)} dx + c\]
We know the integral of ${e^{2x}}$ with respect to x is \[\dfrac{{{e^{2x}}}}{2}\].
So, we get,
\[ \Rightarrow f\left( x \right) \times \left( {\dfrac{{{e^{2x}}}}{2}} \right) - \int {\left[ {f'\left( x \right) \times \left( {\dfrac{{{e^{2x}}}}{2}} \right)} \right]dx} + \int {{e^{2x}}f'\left( x \right)} dx + c\]
Simplifying further,
\[ \Rightarrow \left( {\dfrac{{f\left( x \right) \times {e^{2x}}}}{2}} \right) - \dfrac{1}{2}\int {\left[ {f'\left( x \right) \times {e^{2x}}} \right]dx} + \int {{e^{2x}}f'\left( x \right)} dx\]
Adding up like integral terms,
\[ \Rightarrow \left( {\dfrac{{f\left( x \right) \times {e^{2x}}}}{2}} \right) + \dfrac{1}{2}\int {\left[ {f'\left( x \right) \times {e^{2x}}} \right]dx} + c\]
Substituting the value of integral $\int {{e^{2x}}f'\left( x \right)} dx$ as $g\left( x \right)$, we get,
\[ \Rightarrow \left( {\dfrac{{f\left( x \right) \times {e^{2x}}}}{2}} \right) + \dfrac{1}{2}g\left( x \right) + c\]
Taking $\dfrac{1}{2}$ common from both terms,
\[ \Rightarrow \dfrac{1}{2}\left[ {f\left( x \right){e^{2x}} + g\left( x \right)} \right] + c\], where c is the arbitrary constant.

Hence, option (B) is the correct answer.

Note:
Integration by parts method can be used to solve integrals of various complex functions involving the product of functions within itself easily. In the given question, the use of integration by parts method once has made the process much easier and organized. However, this might not be the case every time. We may have to apply the integration by parts method multiple times in order to reach the final answer.