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

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. $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x} \cdot \dfrac{{dx}}{{dx}} = {e^x}$.