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

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so
so
<math>y=\int\limits_{1-3x}^{1}\frac{1}{(1+u^2)}x^3, dx</math>
<math>y=\int\limits_{1-3x}^{1}\frac{1}{(1+u^2)}x^3, dx</math>
=
=
<math>(0)*f(1)-(-3)*f(1-3x)</math>
<math>(0)*f(1)-(-3)*f(1-3x)</math>

Revision as of 01:31, 24 August 2022

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(x)}

so 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 y=\int\limits_{1-3x}^{1}\frac{1}{(1+u^2)}x^3, dx}

= 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)}