Question

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

Answer 1

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.