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anonymous
 one year ago
Need Help! Use the definition of an integral to evaluate the integral: 5 to 1, (1+3x)*dx
anonymous
 one year ago
Need Help! Use the definition of an integral to evaluate the integral: 5 to 1, (1+3x)*dx

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_{1}^5(1+3x)dx\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you have to do it the hard way right? not the easy way ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439170371150:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea, thats it! and the hard way.... and sorry about the drawing above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first one is evidently a constant since \[\sum_{i=1}^n 1=n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep! I dont know what to do about the negative 3 though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the first part \[\int_{1}^51dx=6\] which is not surprising since \(1\times 6=6\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't get the negative 3 part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_{1}^53xdx=3\int_{1}^5xdx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so pull the 3 outside the summation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's supposed to be a negative 3 though...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean not \[\int_{1}^5(1+3x)dx\] that is a plus 3, is there a typo there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then there is no \(3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which type of way are you solving for this problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks like same as you break the integral in to two pieces, first one is \[\int_{1}^51dx=\frac{6}{n}\sum_{1}^n1=\frac{6}{n}\times n=6\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have that part above right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok that might work too, although it is certainly more work than necessary

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea, but that's the way we gotta do it on our exam

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0really? why can't you just pull the 3 out of the summation like a normal person would?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You mean make the 6/n that's already on the outside and multiply that with negative 3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no matter lets see if this works, it should you have \\[\frac{6}{n}\sum_{i=1}^n(3)=\frac{6}{n}\times 3n=18\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that totally clear or no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0eh...actually how did you get the n onto the negative 3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so far we are at \(618=12\) now for the rest (i hope this wrods)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok lets go real slow here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first sum is what you wrote above \[\frac{6}{n}\sum_{i=1}^n1\]right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now what does \[\sum_{i=1}^n1\] mean in reality?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right it is \[\overbrace{1+1+...+1}^{\text{ n times}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0similarly \[\sum_{i=1}^n c=n\times c\] for any constant \(c\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay well that makes everything more clear

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \[\sum_{i=1}^n(3)=3\sum_{i=1}^n1=3n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is just the distributive law in action, which is why you should really start with \[\int 1dx+3\int xdx\] to make your summations simpler i don't get the "do it this way on a test" part, but i believe you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now we have taken care of the first two sums correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=3x+1\] is a line with slope 3 and y intercept 1 it would be easiest to compute this integral with geometry lets continue

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the answer was 42 in the back of the book

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all we have left is \[\frac{6}{n}\frac{18}{n}\sum_{i=1}^ni\] right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the last summation after you distribute the \(\frac{6}{n}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you made some sort of goofy algebra mistake in the last line should have 3 sums, not a product with 3 in it but you wrote the sum correctly \[\sum_{i=1}^ni=\frac{n(n+1)}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now we are at \[\frac{6\times 18}{n^2}\times\frac{n(n+1)}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, have that already

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or more precisely \[\lim_{n\to \infty}\frac{6\times 18}{n^2}\times\frac{n(n+1)}{2}\] a limit you compute with your eyeballs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it clear that the limit is 54?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just made some silly mistake and yea

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol thanks! Appreciate it the help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i guess \[618+54=40\] or sommat

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw it really really is easier if you pull the 3 out of the summation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0didn't you say it was 40 in the back of the book?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0....no lol, scroll up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0who don't know that?
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