anonymous
  • anonymous
Determine if convergent or divergent. If convergent, where does it converge to? Integral from -infinte to 0 of 2^r.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{-\infty}^{0}2^r\,dr\,\,\,\text{ Is this what you have?}\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
so 2^r=y... ln y = rln 2=> y = e^(rln 2). let u = rln 2, du = ln 2 then \[\int\limits_{-\infty}^{0}2^{r}\,dr=\frac{1}{\ln 2}\int\limits_{-\infty}^{0}(e^{r\ln 2})(\ln 2)\,dr=\frac{1}{\ln 2}\int\limits_{a}^{b}e^{u}\,du\] can you take it from here?

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anonymous
  • anonymous
Are you sure du = ln2? wouldn't you use product rule?
anonymous
  • anonymous
wait nvm ln2 is constant
anonymous
  • anonymous
du = ln 2 dr... sorry, it must have fell off
anonymous
  • anonymous
and yes... ln 2 is constant
anonymous
  • anonymous
it's almost 1
anonymous
  • anonymous
so... do you follow it? does it make sense? can you finish from here or do you still need some assistance?
anonymous
  • anonymous
How do i evaluate is though? I get 1/ln2 - (e^-inf)/ln2
anonymous
  • anonymous
okay... just a sec
anonymous
  • anonymous
\[\frac{1}{\ln 2}\int\limits\limits_{a}^{b}e^{u} \,du = \frac{e^{u}}{\ln 2}\huge{|_{\normalsize{a}}^{\normalsize{b}}}\] \[= \frac{e^{r\ln{2}}}{\ln 2}\huge{|_{\normalsize{-\infty}}^{\normalsize{0}}} = \normalsize \frac{2^{r}}{\ln 2}\huge{|_{\normalsize{-\infty}}^{\normalsize{0}}} = \normalsize\frac{1}{\ln{2}}- \lim_{r \rightarrow -\infty}\frac{2^{r}}{\ln{2}}\]
anonymous
  • anonymous
whew... that took a bit to figure out how to type!
anonymous
  • anonymous
sorry for the big =.
anonymous
  • anonymous
thank you
anonymous
  • anonymous
make sense? the limit should go to 0 so the integral converges... right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
okay good... always when you have a base other than e, to take a derivative or integrate, you'll always change to a base of e and then manipulate as needed. keep that in mind.

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