5.3 The Fundamental Theorem of Calculus/17

FTC #1

$G(x)=f^{\prime }(x)$ or in other words ${\frac {d}{dx}}[\int \limits _{a(x)}^{b(x)}F(x)dx]$ is $\ b^{\prime }(x)*f(b(x))-a^{\prime }(x)*f(a(x))$

$y=\int \limits _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du$

so $y=\int \limits _{1-3x}^{1}{\frac {1}{(1+u^{2})}}u^{3},du$

using the formula we get y=$(0)*f(1)-(-3)*f(1-3x)$

which is equal to $(3)*f(1-3x)$

which is=$3*(1-3x)^{3}*{\frac {1}{(1+(1-3x)^{2})}}$

or simplified to ${\frac {3*(1-3x)^{3}}{(1+(1-3x)^{2})}}$

FTC #2 (not done yet)

$\int \limits _{a}^{b}f(x)dx$ is equal to $F(b)-F(a)$ Where F is the antiderivative of f such that $F^{\prime }=f$

$y=\int \limits _{1-3x}^{1}{\frac {x^{3}}{(1+u^{2})}}dx$

$y=\int \limits _{1-3x}^{1}{\frac {x^{3}}{(1+u^{2})}}dx$

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