Integral of $$$\frac{\sqrt{11} e^{- \frac{x}{2}}}{22}$$$
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Find $$$\int \frac{\sqrt{11} e^{- \frac{x}{2}}}{22}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{\sqrt{11}}{22}$$$ and $$$f{\left(x \right)} = e^{- \frac{x}{2}}$$$:
$${\color{red}{\int{\frac{\sqrt{11} e^{- \frac{x}{2}}}{22} d x}}} = {\color{red}{\left(\frac{\sqrt{11} \int{e^{- \frac{x}{2}} d x}}{22}\right)}}$$
Let $$$u=- \frac{x}{2}$$$.
Then $$$du=\left(- \frac{x}{2}\right)^{\prime }dx = - \frac{dx}{2}$$$ (steps can be seen »), and we have that $$$dx = - 2 du$$$.
Therefore,
$$\frac{\sqrt{11} {\color{red}{\int{e^{- \frac{x}{2}} d x}}}}{22} = \frac{\sqrt{11} {\color{red}{\int{\left(- 2 e^{u}\right)d u}}}}{22}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-2$$$ and $$$f{\left(u \right)} = e^{u}$$$:
$$\frac{\sqrt{11} {\color{red}{\int{\left(- 2 e^{u}\right)d u}}}}{22} = \frac{\sqrt{11} {\color{red}{\left(- 2 \int{e^{u} d u}\right)}}}{22}$$
The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:
$$- \frac{\sqrt{11} {\color{red}{\int{e^{u} d u}}}}{11} = - \frac{\sqrt{11} {\color{red}{e^{u}}}}{11}$$
Recall that $$$u=- \frac{x}{2}$$$:
$$- \frac{\sqrt{11} e^{{\color{red}{u}}}}{11} = - \frac{\sqrt{11} e^{{\color{red}{\left(- \frac{x}{2}\right)}}}}{11}$$
Therefore,
$$\int{\frac{\sqrt{11} e^{- \frac{x}{2}}}{22} d x} = - \frac{\sqrt{11} e^{- \frac{x}{2}}}{11}$$
Add the constant of integration:
$$\int{\frac{\sqrt{11} e^{- \frac{x}{2}}}{22} d x} = - \frac{\sqrt{11} e^{- \frac{x}{2}}}{11}+C$$
Answer
$$$\int \frac{\sqrt{11} e^{- \frac{x}{2}}}{22}\, dx = - \frac{\sqrt{11} e^{- \frac{x}{2}}}{11} + C$$$A