Finding the Derivative of an Integral with Limited Domain

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Homework Help Overview

The discussion revolves around finding the derivative of a function defined as an integral, specifically f(x) = ∫e^sin(t) dt, with limits involving trigonometric and polynomial expressions. Participants explore the application of the Fundamental Theorem of Calculus in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants question the conditions under which the Fundamental Theorem of Calculus applies, particularly regarding the continuity of the integrand and the definition of the sine function. There is confusion about the limits of integration and how they relate to the derivative.

Discussion Status

Some participants have provided guidance on applying the Fundamental Theorem of Calculus, suggesting that the derivative can be found using the formula for derivatives of integrals with variable limits. However, there remains some uncertainty about the specific setup of the integral and its implications.

Contextual Notes

There is mention of confusion regarding the limits of integration and their relevance to the problem, as well as concerns about the continuity of the sine function within the specified domain.

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Homework Statement


Find the derivative of the function:

[itex]f(x)= ∫e^sin(t) dt.[/itex] (A is cos (x) and B is (x^2))

Homework Equations


The Attempt at a Solution



I read this site http://mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html and I was wondering how I would be able to determine what problems this technique would work for. It says it has to be an open interval, and the function must be continuous. But in this case, since sin is defined only from -1 to 1, this would not work right? Then what must I do in such a case? Do I take the integral of the function and then plug in the upper and lower limits? I am utterly confused. Thank you.
 
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Justabeginner said:

Homework Statement


Find the derivative of the function:

[itex]f(x)= ∫e^sin(t) dt.[/itex] (A is cos (x) and B is (x^2))
You mean [itex]f(x)= \int e^{sin(t)}dt[/itex]. But what do "A" and "B" have to do with this?
And how does "x" come into it?

If you mean [itex]f(x)= \int_a^x e^{sin(t)} dt[/itex] then the derivative with respect to x is given by the "Fundamental Theorem of Calculus": The derivative, with respect to x, of [itex]\int_a^x F(t)dt[/itex] is F(x) no matter what a is.

Homework Equations


The Attempt at a Solution



I read this site http://mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html and I was wondering how I would be able to determine what problems this technique would work for. It says it has to be an open interval, and the function must be continuous. But in this case, since sin is defined only from -1 to 1, this would not work right? Then what must I do in such a case? Do I take the integral of the function and then plug in the upper and lower limits? I am utterly confused. Thank you.
 
A and B are the lower and upper limits of the integral but every time I put it in LaTex form it didn't pull up as I wanted it to, so I took it out. X is the variable used in the lower and upper limits of the integral.

So the integral limits are irrelevant to the problem itself, and by the Fundamental Theorem of Calculus, the derivative of ANY integral is F(x)?

Thank you.
 
He means$$
f(x) = \int_{\cos x}^{x^2}e^{\sin t}\, dt$$
 
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This will allow you to quickly apply the fundamental theorem :

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))(b'(x)) - f(a(x))(a'(x))$$
 
So in any question which asks me to find the derivative of an integral, this formula would be utilized?

[2x* e^ sin(x^2)] + [e^(sin(cos x)) * sin x] -Is this a correct application of the rule?

Thank you.
 
Yes, the formula posted by Zondrina is generally applicable. Furthermore, you correctly applied it to your problem. Good work! :-)
 
Thank you! :)
 

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