Question

definite integral help

Answer 1

?

Answer 2

There's an attachment in the other posts mentioning this problem.

Answer 3

Oh, yeah.

Answer 4

Integrate term by term

Answer 5

but i need to commput it as the limit of riemman suns

Answer 6

oh bother.

Answer 7

as the norm partition goes to zero

Answer 8

i just need somone to help me solve this one problem so that i can see how its dome

Answer 9

http://www.twiddla.com/576536

Answer 10

or at least wolk me throught he general step of how to do it

Answer 11

It's been a while since I've done riemann sums, but don't you usually start with it in the reimann sum form, then when you take the limit it becomes the integral? Seems to me you're doing it backward?

Answer 12

now there gave me this intergal

Answer 13

and told me to compute it as the limit of Rimmean sums

Answer 14

Geez, I haven't evaluated a limit like that for a while. Do you need to do it analytically? You could just do it numerically if you keep the definition of a limit in mind.

Answer 15

yeah, but i wanna do this one problem the way they want it done, and then the rest i can just do numerically

Answer 16

anybody can help me out

Answer 17

\[\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\Delta x\]

\[f(x)=3x^2-2x+1\]

\[a=-1,b=2\]

\[\Delta x=\frac{b-a}{n}=\frac{3}{n}\]

you can use

\[x_i^*=a+i\Delta x=-1+i\frac{3}{n}\]

put all together and take limit...really not that bad

Answer 18

yes we can do this . ala zarkon yes?

Answer 19

find

\[\lim_{n\to\infty}\sum_{i=1}^{n}\left[3\left(-1+i\frac{3}{n}\right)^2-2\left(-1+i\frac{3}{n}\right)+1\right]\frac{3}{n}\]

Answer 20

go get em!

Answer 21

you will need to use

\[\sum_{i=1}^{n}1=n\]

\[\sum_{i=1}^{n}i=\frac{n}{n+1}{2}\]

\[\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 22

oops

Answer 23

\[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\]

Answer 24

\[\lim_{n\to\infty}\sum_{i=1}^{n}\left[3\left(-1+i\frac{3}{n}\right)^2-2\left(-1+i\frac{3}{n}\right)+1\right]\frac{3}{n}\]

\[=\frac{3}{n}\lim_{n\to\infty}\sum_{i=1}^{n}\left[3\left((-1)^2+-2i\frac{3}{n}+\left(i\frac{3}{n}\right)^2\right)-2\left(-1+i\frac{3}{n}\right)+1\right]\]

Answer 25

I think you can do the algebra from here...really too messy to keep typing

Answer 26

thanks man i appreaciate it

Answer 27

$$=\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left[3\left((-1)^2+-2i\frac{3}{n}+\left(i\frac{3}{n}\right)^2\right)-2\left(-1+i\frac{3}{n}\right)+1\right]$$

$$=\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left[3-i\frac{18}{n}
+i^2\frac{27}{n^2}+2-i\frac{6}{n}+1\right]$$

$$=\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left[6-i\left[\frac{18}{n}+\frac{6}{n}\right]
+i^2\frac{27}{n^2}\right]$$


$$=\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}6-\frac{24}{n}\frac{3}{n}\lim_{n\to\infty}\sum_{i=1}^{n}i
+\frac{27}{n^2}\frac{3}{n}\lim_{n\to\infty}\sum_{i=1}^{n}i^2$$

$$=\lim_{n\to\infty}\frac{18}{n}\sum_{i=1}^{n}1-\lim_{n\to\infty}\frac{72}{n^2}\sum_{i=1}^{n}i
+\lim_{n\to\infty}\frac{81}{n^3}\sum_{i=1}^{n}i^2$$

$$=\lim_{n\to\infty}\frac{18}{n}n-\lim_{n\to\infty}\frac{72}{n^2}\frac{n(n+1)}{2}
+\lim_{n\to\infty}\frac{81}{n^3}\frac{n(n+1)(2n+1)}{6}$$

$$=18-\frac{72}{2}+\frac{81}{3}=9$$

Let me know if you see any typos ...I already had to fix one ;)