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Math Help - Fundamental Theorm of Calculus

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    Fundamental Theorm of Calculus

    I understand it for the most part and have got the previous problems, but this one is throwing me off quite a bit. The question is to find the derivative using FTC part 1.

    It's the integral of 1-3x to 1 of u^3/(1+u^2) du
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    Hello,
    Quote Originally Posted by sfgiants13 View Post
    I understand it for the most part and have got the previous problems, but this one is throwing me off quite a bit. The question is to find the derivative using FTC part 1.

    It's the integral of 1-3x to 1 of u^3/(1+u^2) du
    You have \int_{1-3x}^1 \frac{u^3}{1+u^2} ~du

    What does FTC part 1 says ?

    F(x)=\int_a^x f(u) ~du ~ \implies ~ F'(x)=f(x)

    So now, here is what I propose to you : arrange the original integral in order to get a smilar form to what the FTC gives.

    F(x)=\int_{1-3x}^1 \frac{u^3}{1+u^2} ~du=-\int_1^{1-3x} \frac{u^3}{1+u^2} ~du (because in the FTC, t is on the upper boundary - and there's a - sign because you inverted the order of boundaries)


    Now, what if you let t=1-3x ? You will be able to let t in the upper boundary of the integral. And x=\frac{1-t}{3}


    Hence F \left(\tfrac{1-t}{3}\right)=-\int_1^t \frac{u^3}{1+u^2} ~du

    Apply FTC :

    \left(F \left(\tfrac{1-t}{3}\right)\right)'=-\left[\frac{t^3}{1+t^2}\right]

    Apply chain rule in the LHS :
    \boxed{-\left[\frac{t^3}{1+t^2}\right]}=\left(F \left(\tfrac{1-t}{3}\right)\right)'=\boxed{\frac{-1}{3} F' \left(\tfrac{1-t}{3}\right)}


    Substitute back t=1-3x :

    -\left[\frac{(1-3x)^3}{1+(1-3x)^2}\right]=\frac{-1}{3} \cdot F'(x)

    \boxed{F'(x)=3 \cdot \frac{(1-3x)^3}{1+(1-3x)^2}}


    Does this look clear enough ?
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