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Mathematics: Post your doubts here!

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Guys, its a question of Integration. I'm having trouble understanding how to solve questions in which 2 points are given and relates a constant "k".

2011-12-30 15.45.42.jpg
 
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Guys, its a question of Integration. I'm having trouble understanding how to solve questions in which 2 points are given and relates a constant "k".

Well the question's pretty easy. First you need to integrate dy/dx=x^2(x-k). After integrating you will get
y=(x^4)/4 - (kx^3)/3 + c. Then put x=2 and y=-2 in the integrated equation and get an equation in terms of 'k' or 'c'. I took it out in terms of 'c' and it came out to be c=(8k-18)/3. Then put x=4 and y=2 in the integrated equation along with c=(8k-18)/3. After equating, you will get k=3 and substitute this value of 'k' into c=(8k-18)/3 to find out the value of 'c'. After equating, you will get c=2. Put the values of 'k' and 'c' in the integrated equation and you will get y=(x^4)/4 - x^3 + 2.
 
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Maximum of Cos is 1 and Minimum of Cos is -1.

You have to form two equations and then solve them simultaneously...

Maximum
10=a-b (1)
10=a-b

Minimum
-2=a-b(-1)
-2=a+b

Solve them simultaneously... You will get the answers...
f(x) = a - b cos x,
a and b are POSITIVE constants.
When do two positive constants give a max. value? Upon addition! To obtain the max. value of the function, you insert the minimum value of cos x since that allows the two +ive constants to be added. So, a + b = 10.
Two +ive constants give a minimum value upon subtraction. So you insert cos x = 1 to obtain an expression where b is subtracted from a. Hence, a - b = - 2.
 
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Which paper is that question from? How can i find the very old past papers of 1990's fdor mechanics 1?
 
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Another Q..
In relation to trapezium rule it is asked to state if its underestimate or overestimate along with a reason..
Wat's the reason that shoud be given?
 
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Another Q..
In relation to trapezium rule it is asked to state if its underestimate or overestimate along with a reason..
Wat's the reason that shoud be given?

It's written in my mathematics book that 'If a graph is bending downwards over the whole interval from a to b, then you can be certain that the trapezium rule will give you an underestimate of the true area. If on the other hand, a graph is bending upwards over the whole interval from a to b, then you can be certain that the trapezium rule will give you an over estimate of the true area'. Therefore, the 'reason' would be upwards or downwards. For e.g, by looking at the diagram you can say that 'It's an underestimate because the graph is bending downwards' or 'It's an overestimate because the graph is bending upwards'. By the way, what does the marking scheme say about this question?
 
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It's written in my mathematics book that 'If a graph is bending downwards over the whole interval from a to b, then you can be certain that the trapezium rule will give you an underestimate of the true area. If on the other hand, a graph is bending upwards over the whole interval from a to b, then you can be certain that the trapezium rule will give you an over estimate of the true area'. Therefore, the 'reason' would be upwards or downwards. For e.g, by looking at the diagram you can say that 'It's an underestimate because the graph is bending downwards' or 'It's an overestimate because the graph is bending upwards'. By the way, what does the marking scheme say about this question?

Marking scheme asks us to justify our statement..no particular reason given though..
 
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Marking scheme asks us to justify our statement..no particular reason given though..

Is this a part of Q5, J04, P2? If yes, then I integrated the given equation from 0 to 2 and the answer that I calculated is 0.59~0.60 while as the one calculated using the trapezium rule is 0.50. This show's that the value calculated by the trapezium rule is an underestimate of the area but I don't think that one would've to do so much just for 1 mark. Or you can rephrase the statement I have mentioned above by writing it as, 'As the graph is bending downwards, the area calculated per interval would be smaller therefore the overall area calculated using the trapezium rule would be lesser than the true value of the area'.
 
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