Write the integral that produces the same value as \[\lim_{n \rightarrow \infty } \sum_{i=1}^{n}(3+i(5/n)^2)(5/n)\]
First, what do you think it would be?
\[\int\limits_{0}^{5}(3+5x)dx\]
I just dont understand what goes where
@agentc0re
Sorry the site went down for me, maybe you too?
Yeah it did for me too
Well lets remember that an integral is the sum between a and b where \[\int\limits_{a}^{b} some\ equation\ here\]
okay yeah i got that. would a equal what i equals or something else?
correct, in this case, a=1 and b=\[\infty \]
Okay. Now what about the equation?
well do you think it would be something different than what it currently is?
if it helps, you can change n to be x so you have a dx at the end of the integral. but a dn would be fine too.
no but for all the answers that are given to me they are all different \[\int\limits_{8}^{3}(x^3)dx\] \[\int\limits_{3}^{0}(x+3)^2dx\] \[\int\limits_{5}^{0}(3+5x)dx\]
Ohh i read this wrong. i was thinking that i was imaginary. im dumb. I'm going to work this out on paper real quick and get back with you, okay?
Okay thank you
Im stumped as my calculus is proving to be a bit rusty. Here is what i do know though. http://www.wolframalpha.com/input/?i=lim+n-%3Einf+%28sum_%28i%3D1%29%5En++%283%2Bi%285%2Fn%29%5E2%29%285%2Fn%29%29+ So that limit = 15. Now you just need to compute your choices and see which one gives you an answer of 15.
well none of them gave me 15
Well damn. Im sorry. Try to ping one of the other helpers. I'm stumped on this one.
Okay don't worry about it
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