y(x) is defined by this equation:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0[/tex]

How do I find the tangent line to y(x) at x=0?

Here's what I tried to do:

[tex]y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}[/tex]

Taking the derivative of both sides of the equation:

[tex]y'(x) = ... = (2x+1)^2 = 4x^2 + 4x + 1[/tex]

[tex]y'(x=0) = 1[/tex]

First, is this method correct?

Second, how do I find y(x=0)? I'm guessing that it's 0, but I'm not sure.

Thanks,

Chen

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# Homework Help: Calculus problem (integrals)

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