5.3 The Fundamental Theorem of Calculus/17: Difference between revisions

Revision as of 20:21, 6 September 2022

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

${\frac {d}{dx}}(g(x))={\frac {d}{dx}}\left(\int \limits _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du\right)=(0)*f(1)-(-3)*f(1-3x)$

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

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

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

@@ Line 1: / Line 1: @@
 <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>\frac{d}{dx}(g(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>
 which is equal to <math>(3)*f(1-3x)</math>