anonymous
  • anonymous
Need Help! Use the definition of an integral to evaluate the integral: 5 to -1, (1+3x)*dx
Calculus1
chestercat
  • chestercat
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anonymous
  • anonymous
\[\int_{-1}^5(1+3x)dx\]?
anonymous
  • anonymous
and you have to do it the hard way right? not the easy way ?
anonymous
  • anonymous
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anonymous
  • anonymous
not bad!
anonymous
  • anonymous
Yea, thats it! and the hard way.... and sorry about the drawing above
anonymous
  • anonymous
the first one is evidently a constant since \[\sum_{i=1}^n 1=n\]
anonymous
  • anonymous
Yep! I dont know what to do about the negative 3 though
anonymous
  • anonymous
so the first part \[\int_{-1}^51dx=6\] which is not surprising since \(1\times 6=6\)
anonymous
  • anonymous
i don't get the negative 3 part
anonymous
  • anonymous
\[\int_{-1}^53xdx=3\int_{-1}^5xdx\]
anonymous
  • anonymous
so pull the 3 outside the summation
anonymous
  • anonymous
It's supposed to be a negative 3 though...
anonymous
  • anonymous
not what you wrote
anonymous
  • anonymous
i mean not \[\int_{-1}^5(1+3x)dx\] that is a plus 3, is there a typo there?
anonymous
  • anonymous
nope
anonymous
  • anonymous
then there is no \(-3\)
anonymous
  • anonymous
which type of way are you solving for this problem?
anonymous
  • anonymous
looks 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
  • anonymous
you have that part above right?
anonymous
  • anonymous
I did it like this
anonymous
  • anonymous
ok that might work too, although it is certainly more work than necessary
anonymous
  • anonymous
Yea, but that's the way we gotta do it on our exam
anonymous
  • anonymous
really? why can't you just pull the 3 out of the summation like a normal person would?
anonymous
  • anonymous
You mean make the 6/n that's already on the outside and multiply that with negative 3?
anonymous
  • anonymous
no 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
  • anonymous
is that totally clear or no?
anonymous
  • anonymous
ah, I see it
anonymous
  • anonymous
eh...actually how did you get the n onto the negative 3?
anonymous
  • anonymous
so far we are at \(6-18=-12\) now for the rest (i hope this wrods)
anonymous
  • anonymous
ok lets go real slow here
anonymous
  • anonymous
lol thanks
anonymous
  • anonymous
the first sum is what you wrote above \[\frac{6}{n}\sum_{i=1}^n1\]right ?
anonymous
  • anonymous
mhm
anonymous
  • anonymous
now what does \[\sum_{i=1}^n1\] mean in reality?
anonymous
  • anonymous
n!
anonymous
  • anonymous
right it is \[\overbrace{1+1+...+1}^{\text{ n times}}\]
anonymous
  • anonymous
i.e \(n\)
anonymous
  • anonymous
similarly \[\sum_{i=1}^n c=n\times c\] for any constant \(c\)
anonymous
  • anonymous
okay well that makes everything more clear
anonymous
  • anonymous
so \[\sum_{i=1}^n(-3)=-3\sum_{i=1}^n1=-3n\]
anonymous
  • anonymous
it 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
  • anonymous
so now we have taken care of the first two sums correct?
anonymous
  • anonymous
is the answer -5/2?
anonymous
  • anonymous
i really doubt it
anonymous
  • anonymous
yea, me too
anonymous
  • anonymous
\[y=3x+1\] is a line with slope 3 and y intercept 1 it would be easiest to compute this integral with geometry lets continue
anonymous
  • anonymous
I think the answer was 42 in the back of the book
anonymous
  • anonymous
all we have left is \[\frac{6}{n}\frac{18}{n}\sum_{i=1}^ni\] right?
anonymous
  • anonymous
yea
anonymous
  • anonymous
the last summation after you distribute the \(\frac{6}{n}\)
anonymous
  • anonymous
you 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
  • anonymous
yep
anonymous
  • anonymous
so now we are at \[\frac{6\times 18}{n^2}\times\frac{n(n+1)}{2}\]
anonymous
  • anonymous
yep, have that already
anonymous
  • anonymous
or 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
  • anonymous
I got the answer!
anonymous
  • anonymous
is it clear that the limit is 54?
anonymous
  • anonymous
yay
anonymous
  • anonymous
I just made some silly mistake and yea
anonymous
  • anonymous
lol thanks! Appreciate it the help
anonymous
  • anonymous
and i guess \[6-18+54=40\] or sommat
anonymous
  • anonymous
it equals 42 buddy
anonymous
  • anonymous
btw it really really is easier if you pull the 3 out of the summation
anonymous
  • anonymous
oops
anonymous
  • anonymous
lol you're fine
anonymous
  • anonymous
didn't you say it was 40 in the back of the book?
anonymous
  • anonymous
....no lol, scroll up
anonymous
  • anonymous
u need some rest
anonymous
  • anonymous
who don't know that?

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