Question

Calculate Rn for f(x) = (64 −x^2) on the interval [−4, 5] and write your answer as a function of n without any summation sign

Answer 1

so i have kinda figured that I have 9/n as my change in x
and my Xi is (-4+i9/n)^2

Answer 2

I cannot seem to pull this problem apart as i do the summations. i think i mess up on the algebra or finding constants as i am doing the summation

Answer 3

Well, are you using right hand side approximation?

Answer 4

yes right hand side approximation

Answer 5

so \[\Large
x_i^* = -1+\frac{9i}{n}
\]

Answer 6

\[
f(x_i^*)=64-\left(-1+\frac{9i}{n}\right)^2
\]

Answer 7

I would start by simplifying \(f(x_i^*)\). What are you getting?

Answer 8

well i used (-4 + 9i/n)^2

Answer 9

oh wait, yeah that's right it's -4 not -1.

Answer 10

so i was getting 16-(72/n)i + (81/n^2) i^2

Answer 11

and that is iinside the function without 64 in there

Answer 12

so i would use n(n+1)/2 for i and then i^2 n(n+1)(2n+1)/6

Answer 13

really i am not sure if i am on the right track with all that

Answer 14

that's correct

Answer 15

so doing the summation of each of the parts i get something like 64n-(and here i think i don't get where to go)

Answer 16

Okay give me a second.

Answer 17

k

Answer 18

The first thing I would do is just substitute it in:\[
\Large
R_n = \sum_{n=1}^{\infty} f(x_i^*)\Delta x =\sum_{n=1}^{\infty} \left[64-\left(16-\frac{72i}{n}+\frac{81i^2}{n^2}\right)\right]\left[\frac{9}{n}\right]
\]

Answer 19

Simplify it, slowly.

Answer 20

yea that is similar to what i got. and then i substitute the parts for i in. but the simplification is killing me

Answer 21

Oops, should be an i, not an n. \[
\Large
\begin{split}
\sum_{i=1}^{n} \left[64-\left(16-\frac{72i}{n}+\frac{81i^2}{n^2}\right)\right]\left[\frac{9}{n}\right] \\
&= \sum_{i=1}^{n} \left[ 48+\frac{72i}{n}-\frac{81i^2}{n^2}\right]\left[\frac{9}{n}\right] \\
&= 9\sum_{i=1}^{n} \left[ \frac{48}{n}+\frac{72i}{n^2}-\frac{81i^2}{n^3}\right] \\
\end{split}
\]

Answer 22

crap, it's not showing up

Answer 23

whoa can't see the whole thing lol

Answer 24

\[ \Large
\sum_{i=1}^{n} \left[64-\left(16-\frac{72i}{n}+\frac{81i^2}{n^2}\right)\right]\left[\frac{9}{n}\right]

\]

Answer 25

\[ \Large
\sum_{i=1}^{n} \left[ 48+\frac{72i}{n}-\frac{81i^2}{n^2}\right]\left[\frac{9}{n}\right]
\]

Answer 26

\[
\Large


9\sum_{i=1}^{n} \left[ \frac{48}{n}+\frac{72i}{n^2}-\frac{81i^2}{n^3}\right]
\]

Answer 27

\[ \Large
9 \left( \sum_{i=1}^{n} \frac{48}{n}+\sum_{i=1}^{n} \frac{72i}{n^2}- \sum_{i=1}^{n} \frac{81i^2}{n^3}
\right)
\]

Answer 28

then i can take out the constants right? and put in the values of i?

Answer 29

Yeah

Answer 30

that is very close to how i started. but i don't think i had the 48/n

Answer 31

\[
\Large
9 \left( \frac{48}{n} \sum_{i=1}^{n} 1+ \frac{72}{n^2}\sum_{i=1}^{n} i- \frac{81i^2}{n^3}\sum_{i=1}^{n} i^2
\right)
\]

Answer 32

1 becomes n .. i becomes n(n+1)/2 and i^2 becomes n(n+1)(2n+1)/6

Answer 33

\[\Large
9 \left( \frac{48}{n} (n) + \frac{72}{n^2}\frac{n(n+1)}{2}- \frac{81}{n^3}\frac{n(n+1)(2n+1)}{6}
\right) \]

Answer 34

yea this is definately where the simplification got me messed up

Answer 35

\[
\Large
9 \left( 48 + 36\frac{n^2+n}{n^2}- 13.5\frac{2n^3 +3n^2+n}{n^3}
\right)
\]

Answer 36

okay i can see how that works there. so i would just cancel out what i can and simplify from that and I should have what Rn is?

Answer 37

Yes

Answer 38

so i should have 9(48+36(1+(1/n))-13.5(2+(3/n)+(1/n^2))

Answer 39

yeah, I would distribute it out, and I'd also change the exponents to be negative.

Answer 40

okay cool! Oh man thanks soo much! I got lost in this one

Answer 41

okay i have to go now but i really appreciate it. Now i can see where i went wrong in the whole problem. and my simplified answer was accepted so I know i got it right. Thanks again