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

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FTC #1- <math>G(x)=f^\prime(x)</math>  or in other words <math>\frac{d}{dx}\left[\int\limits_{a(x)}^{b(x)}F(x)dx\right]</math> is <math>\ b^\prime(x)*f(b(x))-a^\prime(x)*f(a(x))</math>
<math>y=\int_{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>
\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}


which is equal to <math>(3)*f(1-3x)</math>
-(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2}


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>




 
<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