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    (Original post by ETRC)
    you can also say area is the same but under x-axis instead of above it
    or you can say modulus area is the same
    I finished the paper early so during the last few minutes after I triple checked every answer I decided to write the magnitude of the integral of g(x) between 0 and 3 is \frac{15}{2}-8\ln2, the same as the area found in ii)
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    (Original post by iAmanze)
    Let me get this straight.

    Integrals measure the area under a graph right?

    This being said, under the graph (if I recall the question accurately) where were two parts, like 2 triangles.

    So, I did 3 x 1 for the rectangle. Took away the two triangles from it to get my answer.

    I wouldn't even know how to do this differently.

    But yes, I was accurate with my limits so should be able to get it like you said.

    Going over exams after they are finished is a bad idea for me lol. From coming out thinking I have an A*, to thinking I have a high A now, somehow if I was wrong on that, I may have a low A...

    With M1 being a low A/ high B it all rests on the shoulders of C4 to see if I get overall an A* an A or even a damn B! O_O! Hate you C4!
    You only needed to do the rectangle minus one integral if I remember correctly.
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    (Original post by Connorbwfc)
    I think I have got between 59 and 64 on this paper. Is it still possible to get an A*?
    Definitely as long as you smash C4
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    (Original post by RemainSilent)
    what ques was that ? I just found the are of integral with limits 4 and 1 not sure if its the same ques as you tho
    That is the easiest/quickest method, and is also what I did. I believe it's question 8.
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    (Original post by SH0405)
    That is the easiest/quickest method, and is also what I did. I believe it's question 8.
    so you did no take away from rectngle or sumthing? I got the same answer as other ut just minus
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    (Original post by Primus2x)
    I finished the paper early so during the last few minutes after I triple checked every answer I decided to write the magnitude of the integral of g(x) between 0 and 3 is \frac{15}{2}-8\ln2, the same as the area found in ii)
    that should be fine
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    how many mark would i get if i didnt subtract three from the rectangle?
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    (Original post by RemainSilent)
    so you did no take away from rectngle or sumthing? I got the same answer as other ut just minus
    I did the 3*1 rectangle minus the value of the integration between 1 and 4. Thought that was the easiest way to find the shaded area?
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    (Original post by Primus2x)
    I did it like this
    

\arcsin{x}=\arccos{x}

\\

x=\sin\left ( \arccos{x} \right )

\\

x=\sqrt{1-\left ( \cos\left ( \arccos{x} \right ) \right )^2} \because \sin^2 \theta+\cos^2 \theta =1 \Rightarrow \sin \theta=\sqrt{1-\cos^2 \theta}

\\x=\sqrt{1-x^2}

x^2=1-x^2

\\

2x^2=1

\\

x^2=\frac{1}{2}

\\

x=\sqrt{\frac{1}{2}}

\\

\therefore x=\frac{\sqrt{2}}{2} \because \text{rationalise denominator}
    you should get marks for this. hopefully a mathematician marks the papers because there are lots of ways of doing some questions that are correct
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    Assuming marks are only lost completely on Q4, a little in terms of accuracy for the last parts of the section B questions and -3 marks for general section A questions could I still get an A in C3?

    Paper was fine, I made lots of silly mistakes as I prepped for NM till 3AM.
    NM went gracefully tho

    Posted from TSR Mobile
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    (Original post by RemainSilent)
    so you did no take away from rectngle or sumthing? I got the same answer as other ut just minus
    I did RECTANGLE TAKE AWAY INTEGRAL.
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    Question 8. (iv) said 'write down the value of...' for the integral - not the area defined by it. Hence, the answer should be negative.
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    (Original post by SH0405)
    Question 8. (iv) said 'write down the value of...' for the integral - not the area defined by it. Hence, the answer should be negative.
    True, luckily I wrote most of the answers twice to make sure, In question one I wrote the coordinates as expressions involving arctangents, exponentials and square roots as well as writing down the approximate decimal coodinates. And in question 8 I wrote both the integral and the magnitude of the integral
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    I think I might have gotten full marks but I am still worried about question 5 because with all the other questions I could check my answer by using different methods or working backwards (e.g. I checked if my integrals were right by seeing if I got back to the original expressions by differentiating) but question 5 can only be solved by differentiating implicitly ( could not even plot it on my calculator since it only plots Cartesian, polar and parametric).

    Can I still get at least 98% if I do question 5 wrong? (I definitely verified that (1,1) is on the curve so I will get marks for it) I might have written the correct answer of 1/2 but I really don't remember.
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    (Original post by Primus2x)
    I did it like this
    

\arcsin{x}=\arccos{x}

\\

x=\sin\left ( \arccos{x} \right )

\\

x=\sqrt{1-\left ( \cos\left ( \arccos{x} \right ) \right )^2} \because \sin^2 \theta+\cos^2 \theta =1 \Rightarrow \sin \theta=\sqrt{1-\cos^2 \theta}

\\x=\sqrt{1-x^2}

x^2=1-x^2

\\

2x^2=1

\\

x^2=\frac{1}{2}

\\

x=\sqrt{\frac{1}{2}}

\\

\therefore x=\frac{\sqrt{2}}{2} \because \text{rationalise denominator}
    Wow!... I just knew that sin45=cos45= 1/root2.. took me about 10 seconds to answer.
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    With the rectangle area question i used substitution and ended up with a weird answer.. How many marks would i lose.
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    (Original post by Duskstar)
    //wip

    3. \,\, \displaystyle \int^2 _1 x^{3}\ln x \, dx

\mathrm{- integrate by parts:}

\dfrac {dv}{dx} = x^{3}, u = lnx

v = \dfrac {x^4}{4}, \dfrac{du}{dx} = \frac{1}{x}

\mathrm{- follow through, and}

= 16 \ln 2 - \frac{15}{16}
    That answer isn't right, it is 4ln(2) not 16: http://www.emathhelp.net/calculators...1&b=2&steps=on
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    How many marks do you think could be lost for writing that 1/3 + 1 = 2/3 instead of 4/3 by accident in Q2?

    Think I win the award for most stupid marks lost.
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    (Original post by Alex J9)
    That answer isn't right, it is 4ln(2) not 16: http://www.emathhelp.net/calculators...1&b=2&steps=on
    Yeah you're right, thanks - I couldn't be bothered to work it out again lol.
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    (Original post by Sophielouuu)

    Think I win the award for most stupid marks lost.
    I'm sorry but I think I'll be taking that :rolleyes:

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