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D/dt

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Calculate d/dt when a=t^2, b=2 and sqrt(x+1)dx.

My answer turns out to be incorrect despite my checking it several times. I got -2/3t* sqrt(t^2+1)

Can someone explain to me what is wrong with my answer and how to go about obtaining it the right way? Thank you everyone. :-)

  • D/dt -

    f(t) = ∫[2,t^2] √(x+1) dx
    df/dt = √(t^2+1) (2t)

    Recall that d/dt ∫[a,b(t)] f(x) dx = f(b(t)) db/dt

    see wikipedia's article on differentiating under the integral sign.

  • D/dt -

    Thank you for the equation and the useful article. :)

    So this is what I get.

    f(t) dt= 2(0)- [sqrt(t^2+1) * 2t]
    f(t) dt= -sqrt(t^2+1) *2t


    Is this correct? Thank you so much. :-)

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