Question

If \[\int {{e^x}} \left( {f\left( x \right) - f'\left( x \right)} \right)dx = \phi \left( x \right)\], then \[\int {{e^x}} f\left( x \right)dx = \]

Answer 1

Hint:
Here we will use the distributive property and solve the equation. Then we will apply the formula of the integration of \[u \times v\] in the equation. Then by simplifying and solving the equation we will get the value of \[\int {{e^x}} f\left( x \right)dx\].

Complete step by step solution:
Given equation is \[\int {{e^x}} \left( {f\left( x \right) - f'\left( x \right)} \right)dx = \phi \left( x \right)\].
Now we will use the distributive property and write the simplified equation. Therefore, we get
\[ \Rightarrow \int {{e^x}} f\left( x \right)dx - \int {{e^x}} f'\left( x \right)dx = \phi \left( x \right)\]
Now we will apply the basic formula of the integration of \[u \times v\] i.e. \[\int {\left( {u \times v} \right)dx} = u\int {vdx} - \int {\left( {u'\int {vdx} } \right)dx} \] and solve the second term of the equation using this formula where \[u = {e^x},v = f'\left( x \right)\]. Therefore, we get
\[ \Rightarrow \int {{e^x}} f\left( x \right)dx - \left( {{e^x}\int {f'\left( x \right)dx} - \int {\left( {{e^x}\int {f'\left( x \right)dx} } \right)dx} } \right) = \phi \left( x \right)\]
We know that the integration of any differential function is equals to the value of the function i.e. \[\int {f'\left( x \right)dx} = f\left( x \right)\]. Therefore, we get
\[ \Rightarrow \int {{e^x}} f\left( x \right)dx - \left( {{e^x}f\left( x \right) - \int {\left( {{e^x}f\left( x \right)} \right)dx} } \right) = \phi \left( x \right)\]
Now we will simplify and solve the above equation to get the value of \[\int {{e^x}} f\left( x \right)dx\]. Therefore, we get
\[ \Rightarrow \int {{e^x}} f\left( x \right)dx - {e^x}f\left( x \right) + \int {{e^x}} f\left( x \right)dx = \phi \left( x \right)\]
Adding the like terms, we get
\[ \Rightarrow 2\int {{e^x}} f\left( x \right)dx = \phi \left( x \right) + {e^x}f\left( x \right)\]
Dividing both sides by 2, we get
\[ \Rightarrow \int {{e^x}} f\left( x \right)dx = \dfrac{{\phi \left( x \right) + {e^x}f\left( x \right)}}{2}\]
Rewriting the above equation, we get
\[ \Rightarrow \int {{e^x}} f\left( x \right)dx = \dfrac{1}{2}\left( {\phi \left( x \right) + {e^x}f\left( x \right)} \right)\]

Hence the value of \[\int {{e^x}} f\left( x \right)dx\] is equal to \[\dfrac{1}{2}\left( {\phi \left( x \right) + {e^x}f\left( x \right)} \right)\].

Note:
Here we have to simplify and solve the equation accordingly so that we will get the value of the desired term. The formula of the integration of \[u \times v\] must be applied only to the second term of the equation because it contains the differential of the function \[f\left( x \right)\] so that when we solve the equation we will get the final equation in terms of \[\int {{e^x}} f\left( x \right)dx\] because it contains the differential of the function \[f\left( x \right)\]. We should know the basic formula of the integration of \[u \times v\] and \[\dfrac{u}{v}\] as well as the formulas of the differentiation of \[u \times v\] and \[\dfrac{u}{v}\].
\[\begin{array}{l}\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\\\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\end{array}\]
Differentiation of the exponential function is equal to itself along with the differentiation of its exponent i.e. \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x} \cdot \dfrac{{dx}}{{dx}} = {e^x}\].