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whats the integral of xe^(4-x) as x approaches infinity?
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\[\int_{}^{\infty} xe^{4-x}dx\] What is the lower limit? really can't do much without that.
oops, sorry! 4
\[\int_{4}^{\infty} xe^{4-x}dx\] is that the problem?
yes correct
its not infinity or -infinity, but I cannot find out the answer
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Integrate by parts u=x and dv=e^4-x change infinity to t, then evaluate as a limit
(as t goes to infinity)
\[\lim_{4 \rightarrow \infty} -x e^{4-x} -e^{4-x} ?\]
? i have no idea...
x--->infinity
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Yep, you got the integral right. now sub in your upper and lower limits. Now you should have something with t in it ( y(t) ) Take the limit of this as t goes to infinity
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