- #1
Waffle07
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Need Help with Solving an Integral Problem? Let Our Experts Guide You!
- Context: Undergrad
- Thread starter Waffle07
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Discussion Overview
The discussion revolves around solving an integral problem involving the application of the fundamental theorem of calculus and the chain rule. Participants seek to clarify their understanding and approach to the problem, which appears to involve a substitution due to the upper limit being a function of sine.
Discussion Character
- Homework-related
- Technical explanation
- Debate/contested
Main Points Raised
- One participant expresses difficulty in applying the fundamental theorem of calculus and the chain rule to their integral problem.
- Another participant emphasizes the importance of defining the functions involved, suggesting that careful application of definitions is necessary.
- A participant proposes a substitution to correct the upper limit of integration, indicating that it is not simply a matter of plugging in values.
- One participant shares a potential answer but acknowledges that it is incorrect, indicating frustration with the problem-solving process.
- A later reply suggests a method for differentiation using the chain rule, providing a specific formulation involving the integral.
- Another participant mentions a resource for additional help, indicating the availability of tutoring services.
Areas of Agreement / Disagreement
Participants do not reach a consensus on the solution to the integral problem, and multiple approaches and interpretations of the fundamental theorem of calculus are presented. The discussion remains unresolved regarding the correct application of the theorem and the resulting answer.
Contextual Notes
Some participants express uncertainty about the definitions of the functions involved and the application of the chain rule. There are references to specific mathematical steps that are not fully resolved, indicating potential gaps in understanding.
- #2
StatusX
Homework Helper
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What does the fundamental theorem of calculus say? Also, you'll need to use the chain rule.
- #3
Waffle07
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Its says the derivative of g(x) is equal f(x), I know the Fundamental Theorem of Calculus, but I just don't know how to use it alongside the chain rule. Our book only has one example of it, and it doesn't explain it very well.
I think the answer is 5x^4*cos(x^5)*cos(sin(x)^5)+sin(x) but, I know its wrong cus the website won't take it.
I think the answer is 5x^4*cos(x^5)*cos(sin(x)^5)+sin(x) but, I know its wrong cus the website won't take it.
Last edited:
- #4
StatusX
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Yea, but that's meaningless unless you define f(x) and g(x). I'm sure you do know it, but all this problem consists of is carefully applying the definitions, so you should be explicit.
- #5
HallsofIvy
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No matter what f and g are? What a remarkable theorem!waffle07 said:
One form of the fundamental theorem of calculus says that if
g(x)= \int_{x_0}^x f(t)dt
then the derivative of g(x) if f(x). Here your only "problem" is that the upper limit is sin(x) instead of x. Can you make a substitution to correct that? As StatusX says, use the chain rule.
- #6
Waffle07
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Alright that didnt help me at all. Um, let's see can you just tell me the answer?? I think I would be able to see what I am doing wrong if I knew the answer.
- #7
StatusX
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No. Keep trying, or ask more questions.
- #8
Waffle07
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K the answer is cosx*cos(sin(x)^5)+cosx*sinx
- #9
Pseudo Statistic
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An awesome way to think of it is to say,
h(x)=\int_{-5}^{\sin x} f(t) dt = F(\sin x) - F(-5)
You want d/dx of that, so use the chain rule straight off...
\dfrac{d}{dx}(F(\sin x)) = \cos x (f(\sin x))
Where f(t) = \cos (t^5) + t
h(x)=\int_{-5}^{\sin x} f(t) dt = F(\sin x) - F(-5)
You want d/dx of that, so use the chain rule straight off...
\dfrac{d}{dx}(F(\sin x)) = \cos x (f(\sin x))
Where f(t) = \cos (t^5) + t
- #10
Georgie Cartier
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heh - just found a great site for this sort of thing the other day. I can't help you but these guys have sure helped me - great physics tutors cheap plus old q & a even cheaper from their library - great for studying.
www.brainmass.com
www.brainmass.com
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