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secondary 4 | A Maths
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pls help find area of shaded region for f,b and qn 3. (Use integration pls) thank you!!!!!
Date Posted: 8 months ago
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Solutions for questions 1f and 2b. For 2b, note that if the you are trying to find the area of a region below the x axis, you need to take the absolute value or negative of the result you get after performing the standard definite integral steps. I have left my answers in exact form, it should be ok to leave in 3sf form as well.
Date Posted:
8 months ago
hi ! thank you so much! but could u please explain how when you integrate in 2b) how you went from -5cosx/2 to -10sinx/2
For integration, I would say just get really familiar with differentiation and then try to reverse engineer a bit. In this case, you want to integrate -5cos(x/2) which means you want to find out what, when differentiated, gives you -5cos(x/2).
Since there is cos(x/2) in the differentiated result, there must be a sin(x/2) in the integrated result. When we differentiate sin(x/2), we get 1/2 * cos(x/2). This looks rather similar to -5cos(x/2) which is what we want to achieve already. We need to multiply by -10 to get from 1/2 * cos(x/2) to -5cos(x/2) so we multiply -10 to sin(x/2). If we were to check back, differentiating -10sin(x/2) indeed gives us the result of -5cos(x/2), so we can write the integral of -5cos(x/2) as -10sin(x/2) + c (but we can drop the arbitrary constant in this case since this is a definite integral).
Keep practising and drilling integration and differentiation and this will soon become very intuitive and you will be able to write it out in 1 step like in my solution!
Since there is cos(x/2) in the differentiated result, there must be a sin(x/2) in the integrated result. When we differentiate sin(x/2), we get 1/2 * cos(x/2). This looks rather similar to -5cos(x/2) which is what we want to achieve already. We need to multiply by -10 to get from 1/2 * cos(x/2) to -5cos(x/2) so we multiply -10 to sin(x/2). If we were to check back, differentiating -10sin(x/2) indeed gives us the result of -5cos(x/2), so we can write the integral of -5cos(x/2) as -10sin(x/2) + c (but we can drop the arbitrary constant in this case since this is a definite integral).
Keep practising and drilling integration and differentiation and this will soon become very intuitive and you will be able to write it out in 1 step like in my solution!
I’m lazy to go and memorise the particular integration rules and formulas, but you can try integrating by applying those integration rules as well. I personally prefer code-switching between differentiation and integration and doing reverse engineering, so that I don’t have to memorise a new set of rules and formulas (on top of the differentiation rules). But it’s all up to you!
Leave a comment if you still need help / clarifications!
Leave a comment if you still need help / clarifications!
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Solution for Question 3. Similarly, one of the two shaded regions is below the x axis so you need to take the absolute of the definite integral. Leave a comment if you still need help / clarifications!
Date Posted:
8 months ago
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