Question

Hello guys, looking for some calculus help on a specific type of rieman problem. I dont have any problems calculating the value as n goes to infinity ussualy, but this question I cant seem to get the right answer. I think it may be because of the square root. I know the answer is around 38, but I cant get it using the limit definition. Any help would be great and thanks in advance, will post pic below.

Answer 1

So yea I know that delta x is 7/n and Xi is 7i/n

Answer 2

Is it a multiple choice Q? If not, how u know the answer is around 38? U may need to show more than what u have posted on the pic...

Reason i said that is because the infinite series has 7/n as a multiplier, so as n approaches infinity, all terms will approach zero. So something seems to be missing here.

Sorry i could be completely wrong though.

Answer 3

Superdave, that isnt correct. there is a way to solve it and it involves the sigma equations such as i^2= n(2n+1)(n+1)/6

Answer 4

I think its just the sqrt confusing me, because ussualy it isnt to hard you just pull on constants and use the sigma equations

Answer 5

@RobertSn thank you for the explanation - i should have noticed when u mentioned "riemann problem" n keep my mouth shut. Hopefully someone else can help u with this! Good luck :)

Answer 6

\[\sum_{k = 1}^{n}\sqrt{49 - 7\frac{ k }{ n }} * \frac{ 7 }{ n} = 7 * \sum_{k = 1}^{n}\sqrt{49 - 7\frac{ k }{ n }} * \frac{ 1 }{ n} = \]

Answer 7

\[7 * \sum_{k = 1}^{n}f(\frac{ k }{ n }) * \frac{ 1 }{ n} \quad where \quad f(x) = \sqrt{49 - 7x }\]

Answer 8

\[= 7 * \int\limits\limits\limits_{0}^{1}\sqrt{49-7x}~dx\]

Answer 9

Divide the closed interval [0,1] into n equal strips. The width of each strip will be 1/n. The kth strip will be at the location x = k/n. The height of this strip evaluated at the upper end of the interval will be f(k/n).
The product f(k/n) * 1/n is the area of the strip.
The Riemann Sum over the interval [0,1] will be the area under the curve f(x), the x-axis, and x= 0 and x = 1. This is same as integrating f(x) from 0 to 1.

Answer 10

oops, I forgot the squaring part inside the radical.
\[= 7 * \int\limits\limits\limits\limits_{0}^{1}\sqrt{49-(7x)^2}~dx\]

Answer 11

Alright il try that thanks for the help!

Answer 12

You are welcome.

Answer 13

And ranga, are you saying to integrate normally? Like use trig sub?

Answer 14

@ranga

Answer 15

Yes.

Answer 16

Oh. I meen it is not supposed to be done like that

Answer 17

I understand the steps you did up to what you have posted, but the questions wants it as the the lim n approaches infinity, because the class is not at the trig sub section yet

Answer 18

something more like factoring out constants and cancelling ns

Answer 19

using the limit definition of the integral

Answer 20

We are given a Riemann summation problem. Riemann summation is dividing an interval into n equal strips, calculating the area of each strip, adding up the total of all areas of the strips and find the sum as n->infinity. This reduces to the area under the curve which can be computed using a definite integral.

Answer 21

Find the sum as n->infinity of the series:\[\sqrt{49 - (\frac{7}{n})^2} * \frac{7}{n} + \sqrt{49 - (\frac{14}{n})^2} * \frac{7}{n} + ... + \sqrt{49 - (\frac{7n}{n})^2} * \frac{7}{n} \]The kth term of the series is: \[\sqrt{49 - (\frac{7k}{n})^2} * \frac{7}{n} \]The Riemann sum is: \[ \lim_{n \rightarrow \infty} \sum_{k = 1}^{n}\sqrt{49 - (7\frac{ k }{ n })^2} * \frac{ 7 }{ n} = \lim_{n \rightarrow \infty} 7 * \sum_{k = 1}^{n}f(\frac{ k }{ n }) * \frac{ 1 }{ n} \quad where \quad \\f(x) = \sqrt{49 - 49x^2 } = 7 * \sqrt{1 - x^2 } \]\[ \text{The Riemann sum is: } \lim_{n \rightarrow \infty} 7 * \sum_{k = 1}^{n}f(\frac{ k }{ n }) * \frac{ 1 }{ n} = 7 * \int\limits\limits\limits\limits_{0}^{1}7 * \sqrt{1-x^2}~dx \\= 49 * \int\limits\limits\limits\limits_{0}^{1}\sqrt{1-x^2}~dx\]

Answer 22

Yes great, that is more what i was trying to see! thanks

Answer 23

Typing these mathematical symbols is such a hassle. It took less than 5 minutes to solve the problem on paper but a lot longer to convey it through the math editor. :)