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Math Help - Help finding a tangent to integral

  1. #1
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    Help finding a tangent to integral

    given is F(x)= the integral from x=1 to x=2x^3 of the function (3t-4)^2 dt

    I need to find the equation of the tangent line to F(x) at x=-1


    so far i have:

    f=600 (which is the slope of F at x=-1)

    i need help finding the (x,y) coordinates
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  2. #2
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    Quote Originally Posted by gamefreeze View Post
    given is F(x)= the integral from x=1 to x=2x^3 of the function (3t-4)^2 dt

    I need to find the equation of the tangent line to F(x) at x=-1


    so far i have:

    f=600 (which is the slope of F at x=-1)

    i need help finding the (x,y) coordinates
    F(x)=\int_1^{2x^3}(3t-4)^2dt

    Let u=2x^3

    F'(x)=\frac{du}{dx}\frac{d}{du}\int_1^u(3t-4)^2dt

    F'(x)=6x^2[3(2x^3)-4]^2

    F'(-1)=600

    Ok, In get the same thing there. Think of the equation of the tangent line as follows:

    y-F(-1)=600[x-(-1)]

    y-F(-1)=600(x+1)

    So you need to find

    F(-1)=\int_1^{2(-1)^3}(3t-4)^4dt

    So that's just a basic definite integral right? So just evaluate F(-1) by using the substitution rule.
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