Aeron Sim's answer to Chua Xiang Ning's Junior College 1 H2 Maths question.
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Hope this helps. :)
Date Posted:
6 years ago
How do you find 3 for the upper and lower integral?
You see the graph on your GC, you will see that at the lower bound x=2, the graph is in the negative area; f(x) < 0. This means that from x=2 to x=?, I need to integrate -ve f(x) due to the modulus originally.
So this ? Is actually x=3 cos after x=3, the graph is in the positive area already; f(x) > 0. This means that from x=3 to x=p (we know p>3 so p being the upper bound is definitely logically), I need to integrate +ve f(x).
Does my explanation help you understand? Sorry ah I forgot to include an explanation for splitting the modulus into -ve f(x) and +ve f(x).
So this ? Is actually x=3 cos after x=3, the graph is in the positive area already; f(x) > 0. This means that from x=3 to x=p (we know p>3 so p being the upper bound is definitely logically), I need to integrate +ve f(x).
Does my explanation help you understand? Sorry ah I forgot to include an explanation for splitting the modulus into -ve f(x) and +ve f(x).
I get it now. Thanks for explaining
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