Question

Determine if convergent or divergent. If convergent, where does it converge to? Integral from -infinte to 0 of 2^r.

Answer 1

\[\int\limits_{-\infty}^{0}2^r\,dr\,\,\,\text{ Is this what you have?}\]

Answer 2

yes

Answer 3

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?

Answer 4

Are you sure du = ln2? wouldn't you use product rule?

Answer 5

wait nvm ln2 is constant

Answer 6

du = ln 2 dr... sorry, it must have fell off

Answer 7

and yes... ln 2 is constant

Answer 8

it's almost 1

Answer 9

so... do you follow it? does it make sense? can you finish from here or do you still need some assistance?

Answer 10

How do i evaluate is though? I get 1/ln2 - (e^-inf)/ln2

Answer 11

okay... just a sec

Answer 12

\[\frac{1}{\ln 2}\int\limits\limits_{a}^{b}e^{u} \,du = \frac{e^{u}}{\ln 2}\huge{|_{\normalsize{a}}^{\normalsize{b}}}\]
\[= \frac{e^{r\ln{2}}}{\ln 2}\huge{|_{\normalsize{-\infty}}^{\normalsize{0}}} = \normalsize \frac{2^{r}}{\ln 2}\huge{|_{\normalsize{-\infty}}^{\normalsize{0}}} = \normalsize\frac{1}{\ln{2}}- \lim_{r \rightarrow -\infty}\frac{2^{r}}{\ln{2}}\]

Answer 13

whew... that took a bit to figure out how to type!

Answer 14

sorry for the big =.

Answer 15

thank you

Answer 16

make sense? the limit should go to 0 so the integral converges... right?

Answer 17

yes

Answer 18

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.