Basic integrations and evaluations (calculus 2 work)

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In summary, the conversation is about a person asking for help with solving integrals and evaluating them. They share their answers and ask for confirmation if they are correct. The expert summarizes the conversation and advises on possible errors in the answers given. They also provide the correct solutions for two of the integrals.
  • #1
joseph_
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Homework Statement



I have a few questions, and I already have (some of) the right answers (or so I think.. but I don't understand them)

(Just finding the anti-derivative)
1. http://www.codecogs.com/eq.latex?\int_{}^{}\sin\left(8z-5\%20\right)
My answer: -cos (4z^2-5z) (not sure if I have to follow the chain rule in this or not?) If not I think this is right.

(Evaluation)
2. http://www.codecogs.com/eq.latex?\int_{1}^{0}3x^{2}+x-5
My answer: 3.5

When I evaluated this I had
-1 + -1/2 + 5

(Evaluation)
3. http://www.codecogs.com/eq.latex?\int_{-\sqrt[]{3}}^{\sqrt[]{3}}(t+1)(t^{2}+4)
My answer: 29.32
My anti-derivative: 1/4t^4 + 1/3t^3 + 2t^2 + 4t

(Evaluation)
4. http://www.codecogs.com/eq.latex?\int_{1/2}^{1}(1/u^{3}%20-%201/u^{4})
My answer: .833
Antiderivative: -1/2v^-2 - -1/3v^-3


Am I doing these right?
 
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  • #2
Oh, I forgot to mention.

On #2 and #4 someone put them into some computer software and said they came out negative... but I can't find any errors on my work?
 
  • #3
Caveat: I looked at #2 and #4 only.

Your answer for #2 is correct, based on the limits of integration you showed. Your friend might have put the limits of integration into the application in the opposite order, which explains the sign discrepancy.

Your answer for #4 is the right number, but the wrong sign. I get a value of -5/6.
 

1. What is the fundamental theorem of calculus?

The fundamental theorem of calculus states that the derivative and integral are inverse operations of each other. This means that if we take the integral of a function, and then take the derivative of the resulting function, we will end up with the original function.

2. How do I find the area under a curve?

To find the area under a curve, we can use the definite integral. We first need to determine the limits of integration (the points where the curve begins and ends), and then plug these values into the integral. The resulting value will be the area under the curve.

3. What is the difference between a definite and indefinite integral?

A definite integral has specified limits of integration, while an indefinite integral does not. This means that a definite integral will give a specific numerical value, while an indefinite integral will give a general function with a constant of integration.

4. Can I use calculus to find the maximum or minimum of a function?

Yes, we can use calculus to find the maximum or minimum of a function. This is done by taking the derivative of the function and setting it equal to zero. The resulting value(s) will give the x-coordinate(s) of the maximum or minimum point.

5. What are some real-life applications of integration?

Some real-life applications of integration include finding the area under a curve (such as in physics and engineering), calculating volumes and surface areas of objects, and determining the average value of a function over a certain interval. Integration is also used in many economic and financial models to calculate quantities like profit and marginal cost.

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