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100d100t : D20 ama Watch

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    why not eh
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    Why don't you get a bigger sized avatar of the same panda since you're a sub?
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    (Original post by Keoje)
    Why don't you get a bigger sized avatar of the same panda since you're a sub?
    i never knew i could :zomg:
    (tbf i like my panda the way it is :ninjagirl:)
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    What does the title mean? Perhaps rather I could ask you to evaluate \displaystyle \int_0^1 \dfrac{1}{a^{2}+x^{2}} dx

    I might be pretty useless here though
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    (Original post by plusC)
    What does the title mean? Perhaps rather I could ask you to evaluate \displaystyle \int_0^1 \dfrac{1}{a^{2}+x^{2}} dx

    I might be pretty useless here though
    100 Days 100 threads im not sure about the intergration im quite bad at maths
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    (Original post by flou_fboco2)
    100 Days 100 threads im not sure about the intergration im quite bad at maths
    You did say ask me anything
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    (Original post by plusC)
    You did say ask me anything
    touche but i didn't expect maths Qs (my maths teachers would be very disappointed :emo:)
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    (Original post by flou_fboco2)
    touche but i didn't expect maths Qs (my maths teachers would be very disappointed :emo:)
    There there, this is a q towards the end of year 13 maths involving integration by substitution (x=a\tan(\theta)), don't feel bad
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    (Original post by plusC)
    There there, this is a q towards the end of year 13 maths involving integration by substitution (x=a\tan(\theta)), don't feel bad
    :rofl: i just finished y13
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    (Original post by flou_fboco2)
    :rofl: i just finished y13
    Oops :P haha, then maybe you can do it if you've done year 13 maths, nothings impossible
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    (Original post by plusC)
    Oops :P haha, then maybe you can do it if you've done year 13 maths, nothings impossible
    i could had i not forgotten them but i'll try
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    (Original post by flou_fboco2)
    i could had i not forgotten them but i'll try
    Don't forget the substitution, it'll work out nicely then
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    (Original post by plusC)
    Don't forget the substitution, it'll work out nicely then
    okie dokie thanx
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    (Original post by flou_fboco2)
    i never knew i could :zomg:
    (tbf i like my panda the way it is :ninjagirl:)
    Re-upload your panda avatar and it should come up bigger
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    (Original post by flou_fboco2)
    why not eh
    Oh and a question :beard:

    What was the last time you bought?
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    (Original post by plusC)
    Don't forget the substitution, it'll work out nicely then
    i give up :cry2:

    (Original post by Tilly-Elizabeth)
    Oh and a question :beard:

    What was the last time you bought?
    i bought cupcakes:sogood:
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    (Original post by flou_fboco2)
    i give up :cry2:



    i bought cupcakes:sogood:
    That's alright, I don't think we were ever asked something like this in C4. But ill lay out a solution just in case you're interested

    Spoiler:
    Show

    Let I=\displaystyle\int_0^1 \dfrac{1}{a^{2}+x^{2}} dx; x=a\tan(\theta)

    It follows \dfrac{dx}{d\theta}=a\sec^{2}( \theta)

    So dx=a\sec^{2}(\theta)d\theta

    Notice now that x^{2}+a^{2}=(a\tan(\theta))^{2}+  a^{2}=a^{2}(1+\tan^{2}(\theta))=  a^{2}\sec^{2}(\theta) via the identity sec^{2}(\theta)=1+\tan^{2}(\thet  a)

    And the limits, assuming a\not=0

    When x=1, \tan(\theta)=\dfrac{1}{a} -> \theta=\arctan(\frac{1}{a})

    When x=0, tan(\theta)=0 -> \theta=0

    Now finally

    I=\displaystyle \int_{0}^{\arctan(\frac{1}{a})} \dfrac{a\sec^{2}(\theta)}{a^{2}\  sec^{2}(\theta)} d\theta ;\sec^{2}(\theta) cancels from top and bottom leaving us with

    I=\displaystyle \int_{0}^{\arctan(\frac{1}{a})} \dfrac{1}{a} (d\theta) ; since 1/a is a constant, integrating just multiplies 1/a by theta

    I=\left[ \frac{\theta}{a} \right]_{0}^{\arctan(\frac{1}{a})}= \frac{1}{a}\arctan(\frac{1}{a})

    Don't worry I'm not going to scold you for hating maths in the summer haha (even though I should )
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    (Original post by plusC)
    That's alright, I don't think we were ever asked something like this in C4. But ill lay out a solution just in case you're interested

    Spoiler:
    Show

    Let I=\displaystyle\int_0^1 \dfrac{1}{a^{2}+x^{2}} dx; x=a\tan(\theta)

    It follows \dfrac{dx}{d\theta}=a\sec^{2}( \theta)

    So dx=a\sec^{2}(\theta)d\theta

    Notice now that x^{2}+a^{2}=(a\tan(\theta))^{2}+  a^{2}=a^{2}(1+\tan^{2}(\theta))=  a^{2}\sec^{2}(\theta) via the identity sec^{2}(\theta)=1+\tan^{2}(\thet  a)

    And the limits, assuming a\not=0

    When x=1, \tan(\theta)=\dfrac{1}{a} -> \theta=\arctan(\frac{1}{a})

    When x=0, tan(\theta)=0 -> \theta=0

    Now finally

    I=\displaystyle \int_{0}^{\arctan(\frac{1}{a})} \dfrac{a\sec^{2}(\theta)}{a^{2}\  sec^{2}(\theta)} d\theta ;\sec^{2}(\theta) cancels from top and bottom leaving us with

    I=\displaystyle \int_{0}^{\arctan(\frac{1}{a})} \dfrac{1}{a} (d\theta) ; since 1/a is a constant, integrating just multiplies 1/a by theta

    I=\left[ \frac{\theta}{a} \right]_{0}^{\arctan(\frac{1}{a})}= \frac{1}{a}\arctan(\frac{1}{a})

    Don't worry I'm not going to scold you for hating maths in the summer haha (even though I should )
    that's looooong :zomg:
 
 
 
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