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Question Number 66245 by aliesam last updated on 11/Aug/19

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

Commented by Rio Michael last updated on 11/Aug/19

$$lety=e^x$$$$\frac{dy}{dx}=e^x$$$$\int \frac{dy}{dx}=\int e^{x}dx$$$$\Rightarrow y=e^x+c$$

Commented by mathmax by abdo last updated on 11/Aug/19

$$\int e^{x}dx=\int \left(\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\right)dx=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\int x^{n}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{1}{n+1}{x}^{n+1}+c=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}+c$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n!}+c=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}+c-1=e^x+C(C=c-1)$$

Answered by mr W last updated on 12/Aug/19

$$since\frac{d(e^x+c)}{dx}=\frac{d\left(e^x\right)}{dx}+\frac{d\left(c\right)}{dx}=e^x+0=e^x$$$$\Rightarrow \int e^{x}dx=e^x+c$$

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