Integral of $$$e^{32 y}$$$

Find $$$\int e^{32 y}\, dy$$$.

Let $$$u=32 y$$$.

Then $$$du=\left(32 y\right)^{\prime }dy = 32 dy$$$ (steps can be seen »), and we have that $$$dy = \frac{du}{32}$$$.

The integral can be rewritten as

$${\color{red}{\int{e^{32 y} d y}}} = {\color{red}{\int{\frac{e^{u}}{32} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{32}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{e^{u}}{32} d u}}} = {\color{red}{\left(\frac{\int{e^{u} d u}}{32}\right)}}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{32} = \frac{{\color{red}{e^{u}}}}{32}$$

Recall that $$$u=32 y$$$:

$$\frac{e^{{\color{red}{u}}}}{32} = \frac{e^{{\color{red}{\left(32 y\right)}}}}{32}$$

Therefore,

$$\int{e^{32 y} d y} = \frac{e^{32 y}}{32}$$

Add the constant of integration:

$$\int{e^{32 y} d y} = \frac{e^{32 y}}{32}+C$$

$$$\int e^{32 y}\, dy = \frac{e^{32 y}}{32} + C$$$A