Question

What is the integral of \[y - 2y{e^x}\] with respect to \[x?\]

Answer 1

Hint: In order to solve this question, we will use the concept that as we have to find the integration with respect to \[x\] so we will treat \[y\] as a constant term and then integrate the given function using the formulas:
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\]
\[\int {{e^x}dx = {e^x}} + c\]
where \[c\] is the constant of integration.

Complete answer:
We have to find the integral of \[y - 2y{e^x}\] with respect to \[x\]
which means we have to find \[\int {\left( {y - 2y{e^x}} \right)} {\text{ }}dx\]
Let the given integral be \[I\]
i.e., \[I = \int {\left( {y - 2y{e^x}} \right)} {\text{ }}dx{\text{ }} - - - \left( i \right)\]
Now using the difference rule of integration, we know that
\[\int {\left( {f\left( x \right) - g\left( x \right)} \right)} {\text{ }}dx{\text{ }} = {\text{ }}\int {f\left( x \right)} {\text{ }}dx - \int {g\left( x \right)} {\text{ }}dx\]
Therefore, from equation \[\left( i \right)\] we have
\[I = \int {\left( {y - 2y{e^x}} \right)} {\text{ }}dx{\text{ }} = {\text{ }}\int y {\text{ }}dx - \int {2y{e^x}} {\text{ }}dx\]
\[I = {\text{ }}\int y {\text{ }}dx - \int {2y{e^x}} {\text{ }}dx{\text{ }} - - - \left( {ii} \right)\]
Now as we have to find the integration with respect to \[x\] so we will treat \[y\] as a constant term
We know that according to the constant coefficient rule
\[\int {cf\left( x \right)} {\text{ }}dx = c\int {f\left( x \right)dx} \]
Therefore, from equation \[\left( {ii} \right)\] we have
\[I = {\text{ }}y\int {dx} - 2y\int {{e^x}} {\text{ }}dx{\text{ }}\]
\[ \Rightarrow I = {\text{ }}y\int {{x^0}dx} - 2y\int {{e^x}} {\text{ }}dx{\text{ }}\]
Now we know that
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\]
and \[\int {{e^x}dx = {e^x}} + c\]
where \[c\] is the constant of integration.
Therefore, from the above equation, we get
\[I = y\left( {\dfrac{{{x^{0 + 1}}}}{{0 + 1}}} \right) - 2y\left( {{e^x}} \right) + c\]
On solving, we get
\[I = xy - 2y{e^x} + c\]
Hence, the integral of \[y - 2y{e^x}\] with respect to \[x\] is equal to \[xy - 2y{e^x} + c\]

Note: Note that the given integral is an example of an indefinite integral. So, after integrating and finding an indefinite integral, make sure that you add an arbitrary constant \[c\] . Also remember when you are asked to calculate the integral \[f\left( {x,y} \right)\] with respect to \[x\] , \[x\] and \[y\] are assumed as independent variables. The idea behind is that you are summing up the infinitely many \[f\left( {x,y} \right)dx\] as \[x\] goes through all values in the range whereas \[y\] is consistent. Thus, \[x\] is considered to be variable and \[y\] as far as the integral is concerned, considered to be constant. But it does not mean that \[y\] is an unvarying value. It is just that \[y\] is unvarying throughout the infinite sum.