True or False Integral Calculus Question #2

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SUMMARY

The discussion centers on the statement that if $F(x)$ is an antiderivative of a function $f(x)$, then $F(2x)$ is an antiderivative of $f(2x)$. The conclusion drawn is that this statement is false. Through the substitution $u = 2x$, the integral $\int f(2x)dx$ is transformed into $(1/2)F(2x)$, demonstrating that the relationship does not hold as initially proposed. The correct derivative of $F(2x)$ is derived using the chain rule, confirming the confusion surrounding the integration process.

PREREQUISITES
  • Understanding of antiderivatives and the Fundamental Theorem of Calculus
  • Familiarity with substitution methods in integral calculus
  • Knowledge of the chain rule in differentiation
  • Basic concepts of function transformations
NEXT STEPS
  • Study the Fundamental Theorem of Calculus and its implications for antiderivatives
  • Learn more about substitution techniques in integral calculus
  • Explore the chain rule in depth, including its applications in various contexts
  • Practice problems involving transformations of functions in integration
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Students and educators in calculus, particularly those focusing on integration techniques and the properties of antiderivatives. This discussion is beneficial for anyone seeking to clarify concepts related to function transformations and integration methods.

MermaidWonders
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True or False: Let $F(x)$ be an antiderivative of a function $f(x)$. Then, $F(2x)$ is an antiderivative of the function $f(2x)$.
 
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What is the derivative of $F(2x)$? Or, perhaps more to the point, what do you get if you integrate $f(2x)$ using the substitution $u=2x,du=2\,dx$?
 
Kinda confused... how would you do that using substitution?
 
Try it and post your effort. :)
 
OK... Here's what I came up with... took me a while but I'm still kinda confused with myself. :(

Let $u = 2x$. Then $du/dx = 2$ --> $dx = du/2$.

$\int f(2x)dx$ then becomes $\int f(u)(du/2) = (1/2)f(u)du = (1/2)F(u) = (1/2)F(2x)$.
 
So I'm guessing that's false because $\int f(2x)dx \ne F(2x)$ but $(\frac{1}{2})F(2x)$, right?
 
MermaidWonders said:
OK... Here's what I came up with... took me a while but I'm still kinda confused with myself. :(

Let $u = 2x$. Then $du/dx = 2$ --> $dx = du/2$.

$\int f(2x)dx$ then becomes $\int f(u)(du/2) = (1/2)f(u)du = (1/2)F(u) = (1/2)F(2x)$.

That's right. Chain rule. :)

Excellent work!
 
Alright, thanks so much! :)
 

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