Intégrale de $e^{2 x}$

Trouvez $\int e^{2 x}\, dx$.

Let $u=2 x$.

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

Thus,

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

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 $u=2 x$:

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

C'est pourquoi,

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

Ajouter la constante d'intégration :

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

$\int e^{2 x}\, dx = \frac{e^{2 x}}{2} + C$A