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Question Number 63570 by mr W last updated on 05/Jul/19

Commented by mathmax by abdo last updated on 05/Jul/19

$${\int }_0^1{(-1)}^{x}dx={\int }_0^1{e}^{i\pi x}dx={\left[\frac{1}{i\pi }{e}^{i\pi x}\right]}_0^1=\frac{1}{i\pi }\{e^{i\pi }-1\}=\frac{-2}{i\pi }\Rightarrow$$$${\int }_0^1{(-1)}^{x}dx=\frac{2i}{\pi }$$

Commented by mr W last updated on 05/Jul/19

$$thankyousir!$$

Commented by mathmax by abdo last updated on 05/Jul/19

$$youarewelcomesir.$$

Answered by MJS last updated on 05/Jul/19

$$\int a^{x}dx=\frac{a^x}{\ln a}+C$$$$\underset{0}{\stackrel{1}{\int }}{(-1)}^{x}dx={\left[\frac{{(-1)}^x}{\ln -1}\right]}_0^1=-\frac{2}{\ln -1}=\frac{2}{\pi }\mathrm{i}$$

Commented by mr W last updated on 05/Jul/19

$$thankssir!$$

Commented by MJS last updated on 05/Jul/19

$${(-1)}^x=\cos \pi x+\mathrm{i}\sin \pi x$$

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