Finding F'(x) for $\int_{1}^{2x}\frac{1}{1-t^{3}} \,dx$

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Discussion Overview

The discussion revolves around finding the derivative \( F'(x) \) for the integral \( f(x) = \int_{1}^{2x}\frac{1}{1-t^{3}} \,dt \). Participants explore the application of differentiation under the integral sign and the use of substitution in the context of calculus.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant initially presents a formula for \( F'(x) \) but seems to confuse the notation and the integral's variable.
  • Another participant clarifies the notation, suggesting that the correct expression should be \( f(x) = \int_{1}^{2x}\frac{1}{1-t^{3}} \,dt \) and asks for \( f'(x) \).
  • A subsequent reply confirms the clarification and emphasizes that the result should only involve \( x \) and not \( t \).
  • One participant proposes defining \( g(t) = \frac{1}{1-t^{3}} \) and suggests that knowing \( G(t) \) (the antiderivative of \( g(t) \)) allows for the evaluation of \( f(x) \) as \( G(2x) - G(1) \).
  • After taking the derivative, the participant concludes that \( f'(x) = g(2x) \cdot 2 = \frac{2}{1-(2x)^{3}} \), indicating a successful elimination of \( G \) from the expression.
  • Other participants express appreciation for the explanation and the method used to derive the result.

Areas of Agreement / Disagreement

Participants generally agree on the clarification of the notation and the approach to finding the derivative, but there is no explicit consensus on the final expression for \( F'(x) \) as the discussion focuses on the method rather than confirming the correctness of the final result.

Contextual Notes

There is an implicit assumption that participants are familiar with the concepts of differentiation under the integral sign and the properties of antiderivatives. The discussion does not resolve the exact form of \( G(t) \) or its implications for the final expression.

karush
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if $f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dx$ dt, then $F'(x)=$

is this just $\frac{-(2x-1)}{t^2-1}$ ?
 
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karush said:
if $f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dx$ dt, then $F'(x)=$

is this just $\frac{-(2x-1)}{t^2-1}$ ?

Huh? :confused:

Do you perhaps mean:
$$f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dt$$
and
$$f'(x)=?$$
 
I like Serena said:
Do you perhaps mean:
$$f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dt$$
and
$$f'(x)=?$$

yes
 
karush said:
yes

Ah okay. Then the result should contain only $x$ and no $t$.To figure this out let's define $g(t)=\frac{1}{1-t^{3}}$.
And let's suppose we know $G(t)$, so that $G'(t)=g(t)$.

Then:
$$f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dt = \int_{1}^{2x}g(t) \,dt = G(2x) - G(1)$$
We'll leave $G$ like that, since we don't really know $G$.

Taking the derivative, we get:
$$f'(x) = G'(2x) \cdot 2 - 0 = g(2x)\cdot 2 = \frac{1}{1-(2x)^{3}} \cdot 2$$

Hey! We got rid of $G$, so we have the answer! :D
 
Not my question, but I want to say that was a very good excellent explanation. (Smoking)
 
wow that's a cool way to do it(Cool)
 

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