Can you explain how to solve area A? I am confused about the area A and upper area above y=x^2. I thought the area of A should be (2*4)-∫x^2 from 0 to 2
The integral of a curve is the area between the curve and the x-axis. So the integral of (x^2-4) is the area under (x^2-4), which is the area A that I have drawn. Then the area B that I have drawn is the triangular one. The total area that we are looking for is A+B.
Please can you explain why there is not a left quadrant to this sketch?
Hi James — thanks for your question: it’s a very good one. To flesh the details out a bit, what you are basically saying is: “why not consider the quadrant where x < 0 and y > 0?”.
If you do consider this quadrant, you find that the all the inequalities are satisfied by the entire area where x < 0 and y <= 4 and y >= 0, so there is another valid region extending from x = – infinity right up to the point where this area meets the curve y = x^2.
Problem is, if you do this you find that the rest of the question no longer seems to make sense. Firstly, it refers to “the region”, when the answer would now look like two regions; secondly, it asks for the area defined by the inequalities, which would be infinite.
So, when I looked at this question, I concluded that it was not very well worded and it made sense to ignore the left-hand quadrant, because that would have made it impossible to come up with a sensible answer to the whole question. So, as far as I am concerned, I have answered a slightly different, but better worded, question where the word ‘defined’ is replaced by the word ‘bounded’. In this case the infinite region in the x < 0 and y > 0 quadrant is not bounded by all the inequalities, so we do not consider it. And of course if the regions are bounded they are finite and it makes sense to talk about their total area.
Looking back at this answer now, that reasoning still makes sense to me, but maybe I have missed something. What do you think? It wouldn’t be the first time that I had made a stupid mistake, and as a result got totally the wrong end of the stick: if you think I’m talking nonsense, then you’re probably right, so please say.
Thank you very much for your reply!
I had a very similar thought when answering the question and was exercising the possibilities of creating expressions of area in the region where x < 0 and y > 0 in terms of infinity!
Like you say, it is sensible to make the assumption this area is not relevant for the question to make sense, as there is nothing in the three inequalities to suggest we should discount it! However I thought I’d doubt myself before I doubted the wording of the question.
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