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Evaluate the improper integral e^x with limits 0 and -ve infinity.
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\[\int\limits_{-\Pi}^{0}e^x dx\]
1
well the anti-derivative of e^x is e^x then evaluate from -inf to 0 \[= e^{0} - e^{-\infty} = 1 - 0 = 1\]
\[e^{-\infty} = \frac{1}{e^{\infty}} = \frac{1}{\infty} = 0\]
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