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

- anonymous

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

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- anonymous

\[\int_{-1}^5(1+3x)dx\]?

- anonymous

and you have to do it the hard way right? not the easy way ?

- anonymous

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## More answers

- anonymous

not bad!

- anonymous

Yea, thats it! and the hard way.... and sorry about the drawing above

- anonymous

the first one is evidently a constant since
\[\sum_{i=1}^n 1=n\]

- anonymous

Yep! I dont know what to do about the negative 3 though

- anonymous

so the first part
\[\int_{-1}^51dx=6\] which is not surprising since \(1\times 6=6\)

- anonymous

i don't get the negative 3 part

- anonymous

\[\int_{-1}^53xdx=3\int_{-1}^5xdx\]

- anonymous

so pull the 3 outside the summation

- anonymous

It's supposed to be a negative 3 though...

- anonymous

not what you wrote

- anonymous

i mean not
\[\int_{-1}^5(1+3x)dx\] that is a plus 3,
is there a typo there?

- anonymous

nope

- anonymous

then there is no \(-3\)

- anonymous

which type of way are you solving for this problem?

- 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

you have that part above right?

- anonymous

I did it like this

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- anonymous

ok that might work too, although it is certainly more work than necessary

- anonymous

Yea, but that's the way we gotta do it on our exam

- anonymous

really? why can't you just pull the 3 out of the summation like a normal person would?

- anonymous

You mean make the 6/n that's already on the outside and multiply that with negative 3?

- 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

is that totally clear or no?

- anonymous

ah, I see it

- anonymous

eh...actually how did you get the n onto the negative 3?

- anonymous

so far we are at \(6-18=-12\) now for the rest (i hope this wrods)

- anonymous

ok lets go real slow here

- anonymous

lol thanks

- anonymous

the first sum is what you wrote above
\[\frac{6}{n}\sum_{i=1}^n1\]right ?

- anonymous

mhm

- anonymous

now what does \[\sum_{i=1}^n1\] mean in reality?

- anonymous

n!

- anonymous

right it is \[\overbrace{1+1+...+1}^{\text{ n times}}\]

- anonymous

i.e \(n\)

- anonymous

similarly
\[\sum_{i=1}^n c=n\times c\] for any constant \(c\)

- anonymous

okay well that makes everything more clear

- anonymous

so \[\sum_{i=1}^n(-3)=-3\sum_{i=1}^n1=-3n\]

- 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

so now we have taken care of the first two sums correct?

- anonymous

is the answer -5/2?

- anonymous

i really doubt it

- anonymous

yea, me too

- 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

I think the answer was 42 in the back of the book

- anonymous

all we have left is
\[\frac{6}{n}\frac{18}{n}\sum_{i=1}^ni\] right?

- anonymous

yea

- anonymous

the last summation after you distribute the \(\frac{6}{n}\)

- 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

yep

- anonymous

so now we are at
\[\frac{6\times 18}{n^2}\times\frac{n(n+1)}{2}\]

- anonymous

yep, have that already

- 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

I got the answer!

- anonymous

is it clear that the limit is 54?

- anonymous

yay

- anonymous

I just made some silly mistake and yea

- anonymous

lol thanks! Appreciate it the help

- anonymous

and i guess \[6-18+54=40\] or sommat

- anonymous

it equals 42 buddy

- anonymous

btw it really really is easier if you pull the 3 out of the summation

- anonymous

oops

- anonymous

lol you're fine

- anonymous

didn't you say it was 40 in the back of the book?

- anonymous

....no lol, scroll up

- anonymous

u need some rest

- anonymous

who don't know that?

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