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

Revision as of 01:33, 24 August 2022

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

$G(x)=f(x)$

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

= $(0)*f(1)-(-3)*f(1-3x)$ $(3)*f(1-3x)$ $3*(1-3x)^{3}*{\frac {1}{(1+(1-3x)^{2})}}x^{3}+c$

@@ Line 7: / Line 7: @@
 <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>
+<math>(3)*f(1-3x)</math>
+<math>3*(1-3x)^3*\frac{1}{(1+(1-3x)^2)}x^3+c</math>