Evaluating the Derivative of an Integral with Variable Limit

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In summary, the conversation discusses the process of evaluating the derivative of an integral with a variable limit of integration. The fundamental theorem of calculus is used to explain this process, where the derivative of the integral is equal to the integrand evaluated at the upper limit. The conversation also mentions the need to carefully review the proof of the fundamental theorem and the importance of understanding the concept rather than relying on tricks to solve problems.
  • #1
illwerral
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Hi folks! I've taken Calculus I and Calculus II, and I'm honestly not that bad at calculus but there's one thing I never quite got which really troubles me. How does one go about evaluating the derivative of an integral with a variable limit of integration?

Now, I realize that you're supposed to use the fundamental theorem of calculus, and that it somehow works out that, for example:

[tex]
d/dx\int_{a}^{x} 2t dt = 2x
[/tex]

But when I do this, I actually do the integration then do the differentiation... I guess I'm not confident that just replacing t with x (in the example I gave) will work in general, like on a really bad integral like:

[tex]
d/dx\int_{a}^{x} \sqrt{1+t^3}
[/tex]

Does it really equal [tex] \sqrt{1+x^3} [/tex]? I can't actually expand it out to see for sure...

Does this question of mine even make sense or am I crazy? Thanks!
 
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  • #2
[tex]\frac{d}{dx}\int_a^xf(t)dt=\frac{d}{dx}(F(x)-F(a))=f(x)[/tex]
 
  • #3
the only way I can make sense of your question is to translate it like this:

"I like to do things the hard way. How can I do it easily?"

Well, maybe not. That still doesn't make much sense! Have you considered going carefully over the proof of the fundamental theorem?
 
  • #4
Thanks for the replies folks, it's becoming more clear to me. I think I'll have to go over the proof of the fundamental theorem again after it's had time to sink in a bit, but I can solve problems now without feeling as if I'm pulling this out of a bag of tricks!
 

1. What is the purpose of evaluating the derivative of an integral with variable limit?

Evaluating the derivative of an integral with variable limit allows us to find the instantaneous rate of change of a function at a specific point. It can also help us to determine the slope of a tangent line to a curve at a given point.

2. How do you evaluate the derivative of an integral with variable limit?

To evaluate the derivative of an integral with variable limit, we use the Fundamental Theorem of Calculus. This theorem states that if a function f(x) is continuous on an interval [a, b], then the derivative of the integral of f(x) from a to x is equal to f(x).

3. Can the derivative of an integral with variable limit be negative?

Yes, the derivative of an integral with variable limit can be negative. This would indicate that the function is decreasing at that point.

4. Are there any special cases to consider when evaluating the derivative of an integral with variable limit?

Yes, there are two special cases to consider. The first is when the upper and lower limits of the integral are constant values. In this case, the derivative of the integral is simply the derivative of the function evaluated at the upper limit. The second case is when the upper limit is a constant and the lower limit is a variable. In this case, we use the Chain Rule to find the derivative.

5. What is the relationship between the derivative of an integral with variable limit and the original function?

The derivative of an integral with variable limit is simply the original function evaluated at the upper limit. In other words, the derivative and the original function are the same, except the upper limit is now a constant value.

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