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# Another lovely IB problem I dont understand lol

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So yeah this is it, idk where to start

Jan 30, 2021

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If you differentiate g(x)  with respect to x, what do you get?

Jan 30, 2021
edited by Melody  Jan 30, 2021
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This sounds like a fun game!

Guest Jan 30, 2021
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g'(x)

SpaceTsunaml  Jan 30, 2021
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Right so if you integrate g'(x) what do you get?

Melody  Jan 30, 2021
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I understand its finding the area between 2 and 3, and that I know that the points are (2,4), and (3,5), so would the area be 4.5? then -1 so 3.5?

SpaceTsunaml  Jan 30, 2021
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Melody  Jan 30, 2021
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g(x) lol, I feel like im overthinking a lot

SpaceTsunaml  Jan 30, 2021
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Yes you probably are,

Can you answer now or do you want me to keep going?

Actually the integral of  g'(x) is g(x) plus c  but since this is a definite integral you do not need to worry about the c. (it cancels out)

Melody  Jan 30, 2021
edited by Melody  Jan 30, 2021
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Keep going sorry, I already thought of this but my sleep deprived mind has been doing too many of the "Find the volume of a solid by rotating said 2d function around an axis too cope with a simple 3 point problem apparently.

SpaceTsunaml  Jan 30, 2021
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yeah I almost put g(x) + C but thought of that lol

SpaceTsunaml  Jan 30, 2021
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$$\displaystyle \int_2^3(g'(x)-1)dx\\~\\ =\displaystyle \int_2^3(g'(x))dx+\displaystyle \int_2^3(-1)dx\\~\\ =\displaystyle \left[ g(x) \right]_2^3+\displaystyle \left[ -x \right]_2^3$$

you should be able to finish it from there.  Tell me what you get.

Jan 30, 2021
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oh my god, wait it would just be (5-4)-(3-2) right, so just 0?

SpaceTsunaml  Jan 30, 2021
edited by SpaceTsunaml  Jan 30, 2021
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Yes that is right.  (sorry about the confusion)

Melody  Jan 30, 2021
edited by Melody  Jan 30, 2021
edited by Melody  Jan 30, 2021
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ahhhh, okay well my thought process was integral means the g(3) - g(2) + (-3+2). Is this wrong?

oh okayyy you had me freaked out for a sec LOL

SpaceTsunaml  Jan 30, 2021
edited by SpaceTsunaml  Jan 30, 2021