integral of e to the x

proof of the integral of e to the x

$\int e^{x}dx=e^x+c$. We can prove why the integral of $e^x$ is equal to $e^x$ (itself) in several ways, some of which we will show below.

Way 1

$\int e^{x}dx=?$

let $e^x=u$

$d(e^x)=du$ (we take the differential of both sides of the equation)

$(e^x{)}^{\prime }dx=du$

$\boxed{f(x)=e^x\Rightarrow f^{\prime }(x)=e^x}$

$e^{x}dx=du$

$\int e^{x}dx=\int du$ (we take the integral of both sides of the equation)

$\int dx=x+c\Rightarrow \int du=u+c$

$\int e^{x}dx=u+c$

$\int e^{x}dx=e^x+c$

Way 2

$\int e^{x}dx=?$

let $e^x=u$

$ln{e}^x=lnu$

$x.lne=lnu$

$x.1=lnu$

$x=lnu$

$dx=d(lnu)$ (we take the differential of both sides of the equation)

$dx=(lnu{)}^{\prime }du$

$\boxed{f(x)=lng(x)\Rightarrow f^{\prime }(x)=\frac{g^{\prime }(x)}{g(x)}}$

$dx=\frac{u^{\prime }}{u}.du$

$dx=\frac{1}{u}.du$

$dx=\frac{1.du}{u}$

$dx=\frac{du}{u}$

$\int e^{x}dx=\int \cancel{u}.\frac{du}{\cancel{u}}$

$\int e^{x}dx=\int du$

$\int dx=x+c\Rightarrow \int du=u+c$

$\int e^{x}dx=u+c$

$\int e^{x}dx=e^x+c$

Way 3

By using the infinite series expansion of the function $e^x$, we can prove that the integral of $e^x$ is equal to itself. The infinite series expansion of the $e^x$ function is as follows.

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

$\int e^{x}dx=\int (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...)dx$ (we take the integral of both sides of the equation)

$\boxed{\int x^{n}dx=\frac{x^{n+1}}{n+1}+c}$

$\int e^{x}dx=x+\frac{x^2}{2}+\frac{x^3}{3.2!}+\frac{x^4}{4.3!}+\frac{x^5}{5.4!}+...+c$

$\int e^{x}dx=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...+1$ (c=1)

$\int e^{x}dx=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...$

$\int e^{x}dx=e^x$