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