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Improper Integral $$ \int_{-\infty}^0 xe^{2x}dx $$ I have got $$ \lim_{t\to -\infty} [-1/4 - te^{2t} + e^{2t}/4] $$ The answer in My book is $-1/4$ why?

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    $\begingroup$ What happens when you take the limit? $\endgroup$ – Prahlad Vaidyanathan Dec 30 '13 at 8:29
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    $\begingroup$ Let $t = -u$, then $u \to \infty$. Does that help? $\endgroup$ – Hawk Dec 30 '13 at 8:30
  • $\begingroup$ @sidht yes,finally I got it thank you $\endgroup$ – user32104 Dec 30 '13 at 8:35
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Because $$\lim_{t\to-\infty}\left(-\frac{1}{4}-te^{2t}+\frac{e^{2t}}{4}\right)=-\frac{1}{4}$$ where $$\lim_{t\to-\infty}\left(te^{2t}+\frac{e^{2t}}{4}\right)=\lim_{u\to\infty}\left(\frac{-u}{e^{2u}}+\frac{1}{4e^{2u}}\right)=$$ $$=\lim_{u\to\infty}\frac{-u}{e^{2u}}+\lim_{u\to\infty}\frac{1}{4e^{2u}}=0$$ using L'Hopital for first limit $$\lim_{u\to\infty}\frac{-u}{e^{2u}}=\lim_{u\to\infty}\frac{-1}{2e^{2u}}=0$$ for second is clear that it is $0$.

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  • $\begingroup$ Nice to see you on, Adi. :-) $\endgroup$ – mrs Dec 30 '13 at 8:48

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