Integral of $$$2 e^{2 y}$$$

Find $$$\int 2 e^{2 y}\, dy$$$.

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=2$$$ and $$$f{\left(y \right)} = e^{2 y}$$$:

$${\color{red}{\int{2 e^{2 y} d y}}} = {\color{red}{\left(2 \int{e^{2 y} d y}\right)}}$$

Let $$$u=2 y$$$.

Then $$$du=\left(2 y\right)^{\prime }dy = 2 dy$$$ (steps can be seen »), and we have that $$$dy = \frac{du}{2}$$$.

Thus,

$$2 {\color{red}{\int{e^{2 y} d y}}} = 2 {\color{red}{\int{\frac{e^{u}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$2 {\color{red}{\int{\frac{e^{u}}{2} d u}}} = 2 {\color{red}{\left(\frac{\int{e^{u} d u}}{2}\right)}}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$${\color{red}{\int{e^{u} d u}}} = {\color{red}{e^{u}}}$$

Recall that $$$u=2 y$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{\left(2 y\right)}}}$$

Therefore,

$$\int{2 e^{2 y} d y} = e^{2 y}$$

Add the constant of integration:

$$\int{2 e^{2 y} d y} = e^{2 y}+C$$

$$$\int 2 e^{2 y}\, dy = e^{2 y} + C$$$A