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

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


=<math>3*(1-3x)^3*\frac{1}{(1+(1-3x)^2)}x^3+c</math>
=<math>3*(1-3x)^3*\frac{1}{(1+(1-3x)^2)}x^3+c</math>

Revision as of 01:52, 24 August 2022

FTC #1

or in other words Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}} of Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \limits _{a(x)}^{b(x)}F(x)*dx} is Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ b(x)*f(b(x))-a(x)*f(a(x))}

so

using the formula we get y=

=