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exponential growth curve

Please help!!

Determine the area under the following exponential growth curve between 2 seconds and 4
seconds;
integral.PNG
(edited 2 years ago)
Original post by gtdub1234
Please help!!

Determine the area under the following exponential growth curve between 2 seconds and 4
seconds;
𝑦 = 􀶱integral (1 −𝑒^-t) dt


It looks like you need to do a little reading. Do you have a textbook?
Reply 2
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gtdub1234
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i do, where to start!

its the 1-e^-t thats getting me flustered..
Original post by gtdub1234
i do, where to start!

its the 1-e^-t thats getting me flustered..

Can you do ∫241dt\int_2^4 1 dt?

What about ∫24etdt\int_2^4 e^t dt?

What book are you using?

Have you read the relevant chapter and worked through some examples?
(edited 2 years ago)
Reply 4
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gtdub1234
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6
i just panic with math's, its an online HNC edexel , i think maybe i should contact the tutor.

there aren't any examples in the book of this type.

i have probably done questions that use the same method but im unsure on what method to use on this
Reply 5
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gtdub1234
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[ 1e^1 / 1] =[e] ( with 4 & 2 @ top/bottom of brackets)

e^4 - e^2 =

54.6 - 7.39 = 47.21 sq units
(edited 2 years ago)
Original post by gtdub1234
[ 1e^1 / 1] =[e]

e^4 - e^2 =

54.6 - 7.39 = 47.21 sq units

That's correct. I'll send a link for a textbook in a minute.
Reply 7
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gtdub1234
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Original post by Plücker
That's correct. I'll send a link for a textbook in a minute.

cheers for that!
Reply 8
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gtdub1234
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the second part of the question is the same but without the -1 before the e...

do i work it out the same way?
(edited 2 years ago)
Original post by gtdub1234
cheers for that!

https://drive.google.com/drive/folders/0B1ZiqBksUHNYX3dkQXFRQ0NlNDA

Scroll down to the bottom. Look in the year 2 pure book.

Question 5 here is similar. https://activeteach-prod.resource.pearson-intl.com/r00/r0069/r006953/r00695376/current/alevelsb_p2_ex11a.pdf
Reply 10
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gtdub1234
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6
:smile:
Reply 11
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gtdub1234
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Original post by Plücker
That's correct. I'll send a link for a textbook in a minute.

this answer is completely wrong.
i have checked on an integral calculator and on a graphing simulator.
Original post by gtdub1234
this answer is completely wrong.
i have checked on an integral calculator and on a graphing simulator.

Which answer to which question?

I haven't seen an answer to the original question yet. The questions I gave you were intended to find out what you know.

∫24etdt=e4−e2\int_2^4 e^t dt = e^4-e^2 is correct but your original question was ∫24(1−e−t)dt\int_2^4 (1-e^{-t})dt .

You might find it helpful to know that ∫eatdt=1aeat+C\int e^{at} dt = \frac{1}{a} e^{at}+C but you will also eventually (very soon?) need to know some more general principles.
Reply 13
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gtdub1234
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Original post by gtdub1234
Please help!!

Determine the area under the following exponential growth curve between 2 seconds and 4
seconds;
integral.PNG

[ 1e^1 / 1] =[e] ( with 4 & 2 @ top/bottom of brackets)

e^4 - e^2 =

54.6 - 7.39 = 47.21 sq units
Original post by gtdub1234
[ 1e^1 / 1] =[e] ( with 4 & 2 @ top/bottom of brackets)

e^4 - e^2 =

54.6 - 7.39 = 47.21 sq units

That is not the answer to your original question.
Reply 15
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gtdub1234
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Original post by Plücker
That is not the answer to your original question.

thats what i said...
Original post by gtdub1234
thats what i said...

It's wrong because it's the answer to a different question.

This isn't getting you very far.

You could try having a look at the book and the example I posted earlier.

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