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    I don't get the right answer when i do this, and i cant think which bit i am doing wrong.

    a curve has equation y=x^2 arsinhx

    the region R is bounded by the curve the x-axis and the line x=3.

    show that the area of R is 9ln(3+sqrt(10)) - 1/9 (2+7sqrt(10))


    so i integrate by parts, and then use a change of variable to find the second bit. and i thought it was all going pretty well...
    but I got 6ln(3+sqrt(10)) - 2sqrt(10) + 2 (which isnt right)

    any help much much appreciated. xx thanks
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    How have you done the integration by parts?

    If  \int u \frac{dv}{dx} dx= uv- \int v \frac{du}{dx}dx, then we want to get rid of that pesky log (in the arcsinh), so choose

    u=sinh^{-1}x=log(x+\sqrt{x^2+1})

    and then \frac{dv}{dx}=x^2.

    Did you get that, and if you didn't, now can you do it?
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    thanks for replying!!
    yeah, i had that... i think

    ive got:
    [2xarsinhx - integral(2x/sqrt(x^2+1))]

    so the first bit is easy, and the second bit is of the form (f '(x)[f(x)]^n), so thats easy to ingrate too.

    am i on the right track?? im not sure which bit im getting wrong...

    cheers
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    (Original post by rachel123456)
    thanks for replying!!
    yeah, i had that... i think

    ive got:
    [2xarsinhx - integral(2x/sqrt(x^2+1))]

    so the first bit is easy, and the second bit is of the form (f '(x)[f(x)]^n), so thats easy to ingrate too.

    am i on the right track?? im not sure which bit im getting wrong...

    cheers
    If \frac{dv}{dx}=x^2 then what is v? I think that you have v=2x, so you've differentiated where you should have integrated. Try again and see if that works.
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    ooh!!! thanks very much.
    i knew i was doing something stupid. - its worked a charm. cheers!!
 
 
 
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