5.3 The Fundamental Theorem of Calculus/17: Difference between revisions
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<math>g(x)=\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du</math> | |||
<math>g\prime(x)=\frac{d}{dx}\left(\int\limits_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) =(0)*f(1)-(-3)*f(1-3x)</math> | <math>g\prime(x)=\frac{d}{dx}\left(\int\limits_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) =(0)*f(1)-(-3)*f(1-3x)</math> |