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

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


<math>G(x)=f^\prime(x)</math>  or in other words <math>\frac{d}{dx}</math> of <math>\int\limits_{a(x)}^{b(x)}F(x)dx</math> is <math>\ b^\prime(x)*f(b(x))-a^\prime(x)*f(a(x))</math>


<math>y=\int\limits_{1-3x}^{1}\frac{u^3}{(1+u^2)} du</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}


so
-(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2}
<math>y=\int\limits_{1-3x}^{1}\frac{1}{(1+u^2)}u^3, du</math>


using the formula we get y=<math>(0)*f(1)-(-3)*f(1-3x)</math>
</math>


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


which is=<math>3*(1-3x)^3*\frac{1}{(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>
 
FTC #2
 
<math>\int\limits_{a}^{b}f(x)dx</math> is equal to <math>F(b)-F(a)</math> Where F is the antiderivative of f such that <math>F^\prime=f</math>
 
<math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</math>
 
<math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</math>

Latest revision as of 20:31, 6 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=\int _{1-3x}^{1}{\frac {u^{3}}{1+u^{2}}}du}


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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} }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \text{Therefore, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2} }

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