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C2 soloman paper B. integration with an infinate limity

Hey. Could somone help me with question 8(ii)

https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-OCR/Solomon%20C%20QP%20-%20C2%20OCR.pdf

I dont understand how to deal with infinite limits.
thanks

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Original post by SamuelN98
Hey. Could somone help me with question 8(ii)

https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-OCR/Solomon%20C%20QP%20-%20C2%20OCR.pdf

I dont understand how to deal with infinite limits.
thanks


af(x)dx:=limbabf(x)dx\displaystyle\int_a^{\infty} f(x)\mathrm{d}x := \displaystyle\lim_{b\rightarrow \infty} \displaystyle\int_a^b f(x)\mathrm{d} x

The := means defined as. You'd cover this in FP1 and FP3. You probably haven't been taught what a limit is or how to deal with them, but use the intuitive notion of b becoming infinitely large.
Consider first doing the Integral and you would get a result of

2{6x52dx=[4x32]2\displaystyle \int_2^\infty \{\frac{6}{x^\frac{5}{2}} dx = \left[\frac{-4}{x^\frac{3}{2}}\right]_2^\infty

Now, for the infinite integral, replace it with some variable, for example t't' for the meantime and then plug in both values to give f(b)f(a)f(b)-f(a). Now consider taking tt as limt\lim_{t\to \infty}. So that as tt tends to infinity, f(t)f(t) from before would get smaller and smaller so you can consider it as being zero. Therefore you would be left with your result of f(a)f(a) as your final answer.
(edited 7 years ago)
hmmm i don't understand any of this xD but i'd like to understand :/
Reply 4
Original post by thefatone
hmmm i don't understand any of this xD but i'd like to understand :/


What happens to 1x\frac{1}{x} as xx gets really, really, really big? If you're not sure, keep plugging bigger numbers into your calculator, what value does it approach?
Original post by Zacken
What happens to 1x\frac{1}{x} as xx gets really, really, really big? If you're not sure, keep plugging bigger numbers into your calculator, what value does it approach?


it gets infinitely small
Reply 6
Original post by thefatone
it gets infinitely small


And if you had to give a value to it, what value would you give it?
Original post by Zacken
And if you had to give a value to it, what value would you give it?


the infinity symbol??

edit: 0?
Reply 8
Original post by thefatone
the infinity symbol??

edit: 0?


It gets infinitely small and you put the infinity symbol? ._.

0 is correct.

Knowing that, how would you integrate 1dxx2\int_{1}^{\infty} \frac{\mathrm{d}x}{x^2}?
Original post by Zacken
It gets infinitely small and you put the infinity symbol? ._.

0 is correct.

Knowing that, how would you integrate 1dxx2\int_{1}^{\infty} \frac{\mathrm{d}x}{x^2}?


of course xD

hmmm write it without a fraction? i.e dx×x2\mathrm{d}x \times x^{-2} ?
Reply 10
Original post by thefatone
of course xD

hmmm write it without a fraction? i.e dx×x2\mathrm{d}x \times x^{-2} ?


How do you integrate x2dx\int x^{-2} \, \mathrm{d}x?
Original post by thefatone
of course xD

hmmm write it without a fraction? i.e dx×x2\mathrm{d}x \times x^{-2} ?


The notation dxx2\displaystyle\int \dfrac{dx}{x^2} is equivalent to 1x2 dx\displaystyle\int \dfrac{1}{x^2} \ dx.
I suspect this is causing you some confusion . . .
Original post by Zacken
How do you integrate x2dx\int x^{-2} \, \mathrm{d}x?


well raise the power by 1 and divide by the new power

so the result would be x1-x^{-1}


Original post by joostan
The notation dxx2\displaystyle\int \dfrac{dx}{x^2} is equivalent to 1x2 dx\displaystyle\int \dfrac{1}{x^2} \ dx.I suspect this is causing you some confusion . . .
yea just a little xD
Reply 13
Original post by thefatone
well raise the power by 1 and divide by the new power

so the result would be x1-x^{-1}


Yes, so now with limits: [1x]1\displaystyle \bigg[-\frac{1}{x}\bigg]_1^{\infty} - what do you get now?
Original post by SamuelN98
Hey. Could somone help me with question 8(ii)

https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-OCR/Solomon%20C%20QP%20-%20C2%20OCR.pdf

I dont understand how to deal with infinite limits.
thanks


If you just integrate normally, and let X = infinite. Then when you come to your final answer with x's in it. You can take limits. So for example 2/X, as x turns to infitnite, 2/x =0.
Original post by Zacken
Yes, so now with limits: [1x]1\displaystyle \bigg[-\frac{1}{x}\bigg]_1^{\infty} - what do you get now?


that's the bit i was unsure about....
actually

you get [0]-[-1] so the value is just 1?
Reply 16
Original post by thefatone
that's the bit i was unsure about....
actually

you get [0]-[-1] so the value is just 1?


Yeah, that's correct.
Original post by Zacken
Yeah, that's correct.


oh ok thanks a lot ^-^
Reply 18
Original post by thefatone
oh ok thanks a lot ^-^


No problem.
Original post by Zacken
No problem.


so what about OP's question?

i integrated and got 4x324x^ {-\frac {3}{2}}

then i put values in and got

[0]-[1.414]
so is -1.414 correct as an answer?
(edited 7 years ago)

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