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>y=\int_{1-3x}^{1}\frac{u^3}{1+u^2} du</math>




<math>\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}
<math>
\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2}\,du\right) = (0)\cdot\frac{(1)^3}{1+(1)^2}


-(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2}


(0)*f(1)-(-3)*f(1-3x)</math>
</math>


which is equal to <math>(3)*f(1-3x)</math>


which is=<math>3*\frac{(1-3x)^3}{(1+(1-3x)^2)}</math>
<math>
or simplified to <math>\frac{3*(1-3x)^3}{(1+(1-3x)^2)}</math>
\text{Therefore, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2}
 
</math>
 
 
 
 
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Latest revision as of 20:31, 6 September 2022