Question

I need help with determining the definite integral of the following

Answer 1

\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n} \cos((3i/n)+1)(3/n)\]

Answer 2

you the limit?

Answer 3

I know the answer is \[\int\limits_{1}^{4}\cos(x)dx\]

Answer 4

and I know that \[\Delta x = (b - a)/n = (3/n)\]
and
\[x_i = a + i \Delta x\]

Answer 5

and f(x) = cos(x)

Answer 6

but yeah I'm lost anyone able to show me how to solve these type of questions?

Answer 7

I'm not sure srry :(

Answer 8

Oh, this is finding the area under the curve using an infinite limit of a Riemann sum.

Answer 9

I'm I understand the concept I'm just lost when it comes to determing b and a

Answer 10

a and b are just your upper and lower limits of integration.

Answer 11

a is lower; b is upper, I mean.

Answer 12

yes I know

Answer 13

but how did they get 1 to 4

Answer 14

and why is the expression inside tan(x) removed

Answer 15

\[A = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left[ f(c_i) \Delta x \right]\]In your case, the area is \[A = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left[ \cos (\frac {3i}{n}) \times \frac {3}{n} \right]\]You should be able to recognize that f(x) = cos x, Delta x = 3/n and Ci = 3i/n + 1. Now the formula for Ci and Delta x is \[c_i = a + i \Delta x \]\[\Delta x = \frac {b - a}{n}\]Substitute what you have and so you can write Ci as \[\frac {3i}{n} + 1 = a + \frac {3i}{n}\]\[a = 1\]Now b - a = 3 from your delta x. So b - 1 = 3. b = 4. So that limit/sum is the area of cos x from 1 to 4.\[\int\limits_{1}^{4} \cos (x) dx\]

Answer 16

Oops, I made a mistake typing the thing Ci in A = lim... Its\[A = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left[ \cos (\frac {3i}{n} + 1) \times \frac {3}{n} \right]\]

Answer 17

Everything else is correct.

Answer 18

Ok reading it over now

Answer 19

thanks now I see :)

Answer 20

Alright, no problem =)

Answer 21

Hi, I actually have another question that is somewhat harder that I'm confused about

Answer 22

[(4/n)(e^(8i/n) + ln(4i/n))]

Answer 23

what do I do in a case like this?

Answer 24

This? \[A = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left[ \frac {4}{n} ( e^{\frac {8i}{n}} + \ln \frac {4i}{n}) \right]\]

Answer 25

yes

Answer 26

Well, Delta x = 4/n

Answer 27

I have gotten

\[\Delta x = (b - a)/n = 4/n\]

Im confused about finding

\[c_i\]

Answer 28

i Delta x = 4i/n, from that the f(ci) seems like...\[f(c_i) = e^{2c_i} + \ln {ci}\]So \[f(x) = e^{2x} + \ln x\]

Answer 29

oh so that is the function does that render a = 0, and b = 4

Answer 30

Yup, nice =)

Answer 31

So that is \[\int\limits_{0}^{4} (e^{2x} + \ln x )dx\]

Answer 32

thanks :) I think I have a better idea of these now.

Answer 33

the only time I need to worry about whats in the function is when there is a constant

Answer 34

It takes practice, but you'll get it :)