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Intergration

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(edited 5 years ago)

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Original post by Tom123123123
For a positive number a let:
https://imageshack.com/a/img922/2403/LY27l9.png
for what values of a is dI/da=0


What have you tried?
Original post by RDKGames
What have you tried?

integrating by using the exponential rule but that leads nowhere
Original post by Tom123123123
integrating by using the exponential rule but that leads nowhere


You don't really need to explicitly integrate anything.

First part of this kinda important theorem is what you want: https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements
what do mean? show me, please...

Original post by RDKGames
You don't really need to explicitly integrate anything.

First part of this kinda important theorem is what you want: https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements
Original post by Tom123123123
what do mean? show me, please...


Do you not see the relation between FToC and your problem??

D493C8B7-56E2-4427-BA08-CFCAA7B12EF2.jpeg
So F(x)=the integral
and F'(x)=f(x)
and how does that help me?
Original post by Tom123123123
So F(x)=the integral
and F'(x)=f(x)
and how does that help me?


Because in your case, I(a)I(a) is your F(x)F(x) and 42x24-2^{x^2} is your f(t)f(t) ...

The problem can be reformulated using the variables in the theorem;

I(x)=0x42t2.dtI(x) = \displaystyle \int_0^x 4-2^{t^2} .dt

and you seek I(x)I'(x). [Just, yknow, bb in the place of xx]
(edited 5 years ago)
could you just show me, I'm really lost

(so f(t)=F(a)-F(b))
Original post by Tom123123123
could you just show me, I'm really lost

(so f(t)=F(a)-F(b))


Not really sure what is difficult about it?

Using the theorem the answer is I(b)=42b2I'(b) = 4-2^{b^2} to the differentiation bit of the question.
so the answer is ±1
Original post by Tom123123123
so the answer is ±1


Not quite that either.
but I want I'(b) to =0 ?
then b would equal ± square root 2
Original post by Tom123123123
but I want I'(b) to =0 ?


I know... I'm saying those values of bb don't make this hold.

b=±2b=\pm \sqrt{2} as you say above.

Though things to note;

- I forgot the question used aa so I used bb in its place... it's still the same ideas.

- Since the lower limit of the integral is zero, I'd think you disregard the -ve solution. Context is needed tbh.

- Have you even covered FToC? You seem majorly lost on a problem that is trivial given the theorem.
(edited 5 years ago)
Reply 15
Original post by Tom123123123
For a positive number a let:
https://imageshack.com/a/img922/2403/LY27l9.png
for what values of a is dI/da=0

help please

Can I ask what level of maths you're doing e.g. A Level / degree?
further maths- the teacher has not taught us anything :wink:
Reply 17
Original post by Tom123123123
further maths- the teacher has not taught us anything :wink:

Which exam board? And where did you get this question from?
the teacher set the problem....
ocr
Reply 19
Original post by Tom123123123
the teacher set the problem....
ocr

It's a strange question for your teacher to set since this is not part of your spec, although I suppose you can work out the answer if you think about it. Maybe they set it as a challenge?

You don't need to know the theorem posted by RDKGames above for your exam.

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