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secondary 4 | A Maths
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Date Posted: 4 years ago
Views: 211
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This is integration
Thank you
Ok
Has your school touched on integration yet? Or even differentiation? These techniques are needed for these type of questions.
If this is meant to be a coordinate geometry question and you have not learnt differentiation yet, you can skip the area of region III.
I will return to do your vectors question after 12 am if no one has already answered yet.
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Area for part II is 0.5 times (a - 1) times (ae^a - e^a) since (e^a - ae^a) is the y-coordinate of R which is negative, so we must make it positive first.
Area of I is 1/2 x 1 x e^a, not 1/2 x 1 x a, since the y-coordinate of P is e^a, not a.
Area of I is 1/2 x 1 x e^a, not 1/2 x 1 x a, since the y-coordinate of P is e^a, not a.
Thks Bro!For part II,completely forgotten that y-coordinate of R is negative,thks for reminding!As for part I,I have accidentlly taken the x-coordinate of P instead of the y-coordinate of P,thks for pointing out!Cheers!
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Corrected version
Date Posted:
4 years ago
Do U know how to account?
done
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Done. Integration is needed to find the area of region III.
Date Posted:
4 years ago
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