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

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

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

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

So,

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

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 $$$f{\left(u \right)} = e^{u}$$$:

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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