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

Revision as of 02:21, 24 August 2022

FTC #1

$G(x)=f^{\prime }(x)$ or in other words ${\frac {d}{dx}}$ of $\int \limits _{a(x)}^{b(x)}F(x)dx$ 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 $y=\int \limits _{1-3x}^{1}{\frac {1}{(1+u^{2})}}u^{3},du$

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

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

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

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

FTC #2

$\int \limits _{a}^{b}f(x)dx$ is equal to $F(b)-F(a)$ Where F is the antiderivative of f such that $F^{\prime }=f$

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

@@ Line 3: / Line 3: @@
 <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^\prime(x)*f(b(x))-a^\prime(x)*f(a(x))</math>
-<math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</math>
+<math>y=\int\limits_{1-3x}^{1}\frac{u^3}{(1+u^2)} du</math>
 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)}u^3, du</math>
 using the formula we get y=<math>(0)*f(1)-(-3)*f(1-3x)</math>
@@ Line 19: / Line 19: @@
 <math>\int\limits_{a}^{b}f(x)dx</math> is equal to <math>F(b)-F(a)</math> Where F is the antiderivative of f such that <math>F^\prime=f</math>
+<math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</math>
 <math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</math>