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Math 32....how was it? :/

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can anyone solve the volume ques plz??

y = x^(1/2) ln x

Volume = pie ( y )^2
Volume = pie [ x^(1/2) ln x ]^2
Volume = pie [x (ln x)^2]

Now use integration by parts to integrate 'x (ln x)^2'

x (ln x)^2
(x^2/2)(ln x)^2 - [x^2 /2 * 2 ln x * 1/x]
(x^2/2)(ln x)^2 - [x ln x]

Now integrate 'x ln x'

x ln x
(x^2/2) ln x - [x^2/2 * 1 /x]
(x^2/2) ln x - [x/2 ]
(x^2/2) ln x - x^2/4

(x^2/2)(ln x)^2 - [x ln x]
(x^2/2)(ln x)^2 - [ (x^2/2) ln x - x^2/4 ]
(x^2/2)(ln x)^2 - (x^2/2) ln x + x^2/4

Now put the limits 'e' (upper) and '1' (lower).

Volume = pie [(e^2/2)(ln e)^2 - (e^2/2) ln e + e^2/4] - [(1^2/2)(ln 1)^2 - (1^2/2) ln 1 + 1^2/4 ]
Volume = pie [(e^2/2) - (e^2/2) + e^2/4] - [ 1/4 ]
Volume = pie [e^2/4] - [ 1/4 ]
Volume = (pie/4)(e^2 - 1)

Therefore, the volume is '(pie/4)(e^2 - 1)'.
 
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y = x^(1/2) ln x

Volume = pie ( y )^2
Volume = pie [ x^(1/2) ln x ]^2
Volume = pie [x (ln x)^2]

Now use integration by parts to integrate 'x (ln x)^2'

x (ln x)^2
(x^2/2)(ln x)^2 - [x^2 /2 * 2 ln x * 1/x]
(x^2/2)(ln x)^2 - [x ln x]

Now integrate 'x ln x'

x ln x
(x^2/2) ln x - [x^2/2 * 1 /x]
(x^2/2) ln x - [x/2 ]
(x^2/2) ln x - x^2/4

(x^2/2)(ln x)^2 - [x ln x]
(x^2/2)(ln x)^2 - [ (x^2/2) ln x - x^2/4 ]
(x^2/2)(ln x)^2 - (x^2/2) ln x + x^2/4

Now put the limits 'e' (upper) and '1' (lower).

Volume = pie [(e^2/2)(ln e)^2 - (e^2/2) ln e + e^2/4] - [(1^2/2)(ln 1)^2 - (1^2/2) ln 1 + 1^2/4 ]
Volume = pie [(e^2/2) - (e^2/2) + e^2/4] - [ 1/4 ]
Volume = pie [e^2/4] - [ 1/4 ]
Volume = (pie/4)(e^2 - 1)

Therefore, the volume is '(pie/4)(e^2 - 1)'.
i got the formula and the expression right, but messed up with the integration of (ln x)^2. and then completed the ques with the wrong integral. :(
will i get some marks??
 
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That's right! 68 is roughly 91% so its an A* but still you have to score good marks in the other papers. For me I flunked my P1 exam last year but still managed to get a 91% because of my brilliant S1 exam. So A* is actually achieved with both the AS and A2 exams.
im giving accel maths actually p1 il manage 74 and m1 il manage 46 .. s1 is hard for me though :)
 
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If on the differential equation question I integrated the right hand side incorrectly as '2e^2x' instead of '1/2 e^2x' however did all the steps correctly, would I get error carried forward? and if not/so how many marks would I lose?

Jazak Allah Khair.
 
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If on the differential equation question I integrated the right hand side incorrectly as '2e^2x' instead of '1/2 e^2x' however did all the steps correctly, would I get error carried forward? and if not/so how many marks would I lose?

Jazak Allah Khair.

You actually differentiated 'e^2x' instead of integrating it. I have no idea about how many marks you'll get as even the method isn't correct.
 
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Do you remember that question? What was it?

The equation of a curve is y=3sinx+4cos^3x
(i) Find the x-co-ordinates of the stationary points in the interval 0<x<pie. [6]
(ii) Determine the nature of the stationary point in this interval for which x is least. [2]
 
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