Please see the attached image. I do not understand how I am going to go from f(x) to f''(x)....
image?
Wait
There
f(x) = 2 ( x -2) +1 Can you continue to find f'(x) and f"(x)?
i dont think we can just assume that is f, there may be other functions that satisfy those conditions have you tried integration by parts?
How do I know what f is? Sorry I do not know where to start for this one. I am clearly lost for this one.
have you done integration by parts before?
its the reverse of the product rule:\[\int\limits{g'(x).f(x)}dx = f(x)g(x) - \int\limits{f'(x).g(x)}dx\]
Yes
OHHH! So you are suggesting to do integration by parts on the one with the second derivative!
\[\int\limits^2_0{x^2 f''(x)}dx = [x^2f'(x)]^2_0 - \int\limits_0^2{2x f'(x)}dx\]
Wow! It just hit me right now.
something like that :)
OMG! I need to review my integration properties. My final is tomorrow. Thank you for your help. ALSO, I have one more question if you do not mind. I think I can deal with this one now that I know I need to use integration by parts
sure
I just need to attach it, give me a minute
I will message you when I post it.
ok
Ok I posted it.
daamn what level maths is this?
ok dont worry the definition is included
This is Calculus first year.
ah as in a degree? im just curious :)
No this is first year in university ever.
how are you finding it? (dont worry im still looking at your problem)
Math is the hardest. The sciences are pretty straightforward, given you follow the instructor's warnings about the exam.
But this is from my opinion. To me, it looks like it would be easy for you (ie. math)
im fairly sure the lower limit is -5, i just have to go and get a drink, brb
sorry i dont quite understand, are you starting a math degree? what are you studying for? (im sorry if this is me being stupid)
no the lower limit is not -5
i havent seen any maths like this before unfortunatly
The big issue i am having with this problem is the n^2 in the denominator. I have never been exposed to such a case.
ah ok. so can you see that the interval is -5 to 23 ?
im not really using a standard method here as i haven't been taught anything on integration as a limit of a summation yet
consider the first term of that summation, i = 1 we have the term: \[\frac{7 \times14}{n^2}f(-5 + \frac{28}{n})\]
and we know that x* satisfies \[a<x*<a+ \Delta x\] for all n
and\[ x^* = -5 + \frac{28}{n}\] \[a< -5 + \frac{28}{n} < a + \frac{(b-a)}{n}\]
if we take n>0 then the only a which satisfies the first part of the ineq. is a = -5
hmm i think we need \[\le\] in the second part.. im sorry i cant help that much, i havent studied this yet. i will probably cover this stuff next year at university
It is alright. Thanks for your help with the previous question.
no problem :) @experimentX @TuringTest @Foolformath can probably help
???
See attached
Or look at this one:
@satellite73
Any ideas?
Hey try this one I posted just now eigen. Do you know this one?
noo are these riemann sums? i have a calculus book im self teaching but i havent got that far yet
still on its differentiation section
i can see how they are identical, i'm struggling to express it mathematically
its because n tends to infinity, so the first term tends to f(1) in both series, and f(1 + (i-1)/n) tends to f(1 + i/n) for all integer i
(i think)
looks like upper and lower Riemann sums
show they become the same integral then?
any information given on the continuity of function??
For which one experimentX?
Ok let me post out the entire question. Those were only two of the choices given.
c) definite integral gives the area under the curve.
looks like d) the lower and upper Riemann sums also give the area under the curve.
http://pirate.shu.edu/~wachsmut/ira/integ/riemann.html Lemma 7.1.10: Riemann Lemma
How is "a" the same as "b"? I have been trying to get to what they showed in (a) but I wasn't able to. How did they get the one in (a)?
they will not be same ... as n tends to infinity, they will be close enough to each other.
So then why is it that all of the above is the answer?
because that is right ... 1/n will be finitely small for large value of n, similarly, f(1 + i/n) and f(1 + i-1/n) will have close value, so they will be almost same.
if we can determine a number epsilon > 0 such that the difference of two sums is less than epsilon, it proves that they are very close to each other.
Ok I understand. I will keep that in mind. There is one more question, if you are still willing to help. Its alright if you have other things to do.
yeah ... go on
Ok it is this question. Let me just attach it.
Pesky n^2.... >_<
How would I go about with this problem?
somehow looks like a or c, but ... there's stupid front part.
b also fits form the point of view of limit.
i/n = somehow x
That n^2 is throwing me off. I am not used to that. I am just used to seeing: f(xi)*delta x format Where delta x = b-a/n
i/n must be somehow related to x, and 1/n is equivalent to dx
you mean the \(n^2\) in the denominator?
Yes
In the boxed: That is what I am used to seeing:
where delta x = (b-a)/n
7 is giving the trouble ...
i think without 7 it will be equal to b)
one n in the denominator is from \(\Delta x\)
Oo.. still wrong.
the other i/n is related to x
Then what do I do with the other n?
THE OTHER N IS RELATED TO X ALONG WITH i
man now this is bugging me.
|dw:1335573189252:dw| If there were no 14, then b would have been correct.
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