# Where have I gone wrong in my workings ?

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#1

It seems to follow similarly to the mark scheme with us both having the Integrand with a coefficient of 20.

But then we diverge and I arrive at a definite integral which gives me a decimal, unlike the mark-scheme..

Where have I gone wrong ?

Thanks.
Last edited by seals2001; 1 month ago
0
1 month ago
#2
(Original post by seals2001)
It seems to follow similarly to the mark scheme with us both having the Integrand with a coefficient of 20.

But then we diverge and I arrive at a definite integral which gives me a decimal, unlike the mark-scheme..

Where have I gone wrong ?

Thanks.
can we see the question please ?
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#3
(Original post by the bear)
can we see the question please ?
0
1 month ago
#4
(Original post by seals2001)
you went wrong when you tried to integrate Θsin2Θ

if you need to integrate sin2Θ the standard method is to rewrite in terms of cos2Θ and integrate from there.
1
1 month ago
#5
Where you went wrong is the integration of @[email protected] - you would have to do integration by parts again there.
But so much easier to change sin^[email protected] to 1/2-1/[email protected] and integrate that
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#6
(Original post by the bear)
you went wrong when you tried to integrate Θsin2Θ

if you need to integrate sin2Θ the standard method is to rewrite in terms of cos2Θ and integrate from there.
Thank you
1
#7
(Original post by qweqworiet)
Where you went wrong is the integration of @[email protected] - you would have to do integration by parts again there.
But so much easier to change sin^[email protected] to 1/2-1/[email protected] and integrate that
Thank you
0
#8
(Original post by qweqworiet)
Where you went wrong is the integration of @[email protected] - you would have to do integration by parts again there.
But so much easier to change sin^[email protected] to 1/2-1/[email protected] and integrate that
For this question, haven't they used the wrong bounds in the definite integral ?

I.e , shouldn't theta on top be 0 , and theta on the bottom be pi/4. ? NOT the other way around ? The value of theta when it is 0 gives the point further along the x axis.

0
1 month ago
#9
(Original post by seals2001)
For this question, haven't they used the wrong bounds in the definite integral ?

I.e , shouldn't theta on top be 0 , and theta on the bottom be pi/4. ? NOT the other way around ? The value of theta when it is 0 gives the point further along the x axis.
You can swap the limits (but doing so reverses the sign of the integral). You can see this happen between the line starting "Area =" and the next line (in your original post).
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