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

Revision as of 01:50, 24 August 2022

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

$G(x)=f^{\prime }(x)$ or in other words ${\frac {d}{dx}}$ of $\int \limits _{a(x)}^{b(x)}\ F(x)*dx$ is $\ b(x)*f(b(x))-a(x)*f(a(x)$

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 2: / Line 2: @@
 <math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</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(x)*f(b(x))-a(x)*f(a(x) dx</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(x)*f(b(x))-a(x)*f(a(x)</math>
 so