5.3 The Fundamental Theorem of Calculus/17

FTC #1

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 G(x)=f^\prime(x)} or in other words 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}[\int\limits_{a(x)}^{b(x)}F(x)dx]} is $\ b^{\prime }(x)*f(b(x))-a^{\prime }(x)*f(a(x))$

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

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

using the formula we get y=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 (0)*f(1)-(-3)*f(1-3x)}

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

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

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

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