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secondary 4 | A Maths
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Date Posted: 4 years ago
Views: 211
I look at this tonight if no one responds by 4 am. Remind me though, I might forget.
Ok sir. Thank you
Plan changed. Sending you in 2 min.
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done
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This is what you should be considering. Take note that integrals for regions below a curve will give you a negative value, so we must extract its value without the negative sign to talk about its area. Here, I do so by introducing modulus to the entire integral for the part below the x-axis.
Date Posted:
4 years ago
But sir answer is 10/3
Hmm...
I redo later on a separate piece of paper when I return
I redo later on a separate piece of paper when I return
Never mind, I am getting paranoid.
I wrote part a as -1/100 (x - 5)^2 + 1/4 correctly.
And I actually wrote -1/100 (x - 5)^2 + 4 in the second part.
I redo everything for you later.
I wrote part a as -1/100 (x - 5)^2 + 1/4 correctly.
And I actually wrote -1/100 (x - 5)^2 + 4 in the second part.
I redo everything for you later.
Thank you sir. I got it.
Sir , Upper area 5/3
Lower area 5/3
Thank you sir
Lower area 5/3
Thank you sir
I go redo the integration later
Thank you sir
done
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Updated
Date Posted:
4 years ago
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