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    hey guys,
    just cam back from uni. had a test which is like 15% of my cw. pretty confident but just some question that if you can confirm i got right that'll be great.

    1)  8x\sqrt (x^2 + 1) dx integration by substitution between x=o (lower limit) and x=1 (upper limit). Did anyone get 6?

    2) can;t remember exactly by if you have something like  y= \frac{some function}{2} + 2x , to find dy/dx will you just use the quotient rule to find the first part and then "+ 2"?

    thanks
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    I could be wrong, but I think:

    1) Answer = (8/3)*2^(3/2). After integrating, I got (8/3)(x^2+1)^(3/2) -- what did you have?

    2) You could use the quotient rule, but it would be a bit pointless. Just differentiate the numerator, then divide it by 2.
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    (Original post by Prokaryotic_crap)
    hey guys,
    just cam back from uni. had a test which is like 15% of my cw. pretty confident but just some question that if you can confirm i got right that'll be great.

    1)  8x\sqrt (x^2 + 1) dx integration by substitution between x=o (lower limit) and x=1 (upper limit). Did anyone get 6?

    2) can;t remember exactly by if you have something like  y= \frac{some function}{2} + 2x , to find dy/dx will you just use the quotient rule to find the first part and then "+ 2"?

    thanks
    I'm afraid your first answer is wrong. There is a surd in the answer.

    \frac{8(2\sqrt2-1)}{3}

    There is no need to use the quotient rule on the second question (although you should have got the correct answer even if you did).

    \frac{dy}{dx}=\frac{1}{2}\frac{d  (some function)}{dx}+2
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    Oops garbage, damn internet lol. Ignore this post please
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    I got the answer as (16*sqrt(2))/3 - 8/3 for the first one
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     y = 8x \sqrt(x^2 +1) dx             u= x^2+1  \frac{du}{dx} = 2x    dx= \frac{du}{2x} x= 1  u=2  x=0  u=1

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    8(u -1)^\frac{1}{2} u^\frac{1}{2} \frac{du}{2(u-1)^\frac{1}{2}
    simplyfying that and subbung u=2 and u=1 and then taking the latter from the former will give 6 i'm sure. anyone who disagrees please post working. it'll really help.
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    u = sqrt(1+x^2), udu/dx = x, dx = u/x * du substitution into the integral and you get 8u^2 du to integrate and your limits become u=1 to u = sqrt(2)
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    i have just realised my mistake, and to make things worse i have just realised that i had written the right thing and then just crossed it out and written something else. I want to kill myself right now
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    PC,

    This is what it should have looked like.

    \int^1_0 8x \sqrt{x^2 +1} \,dx

    u= x^2+1

    du=2xdx

    4 \int^2_1 u^{\frac{1}{2}} \,du

    I'm sure you can finish it off.
 
 
 
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