X^2+2 were integrated it would be 1/3x^3

  • Thread starter defineNormal
  • Start date
In summary, the conversation is about integrating a function with respect to r and finding the correct constant in front. The person is trying to integrate Ar^(1/2) and is confused about fractions. They mention a general rule for integrating x to a power and ask for clarification on how to apply it to their specific case. They eventually come to the conclusion that the correct answer is (2/3)*Ar^(3/2) +C.
  • #1
defineNormal
10
0
Hello I'm trying to integrate Ar^(1/2)

would the answer be?

well I know if X^2+2 were integrated it would be 1/3x^3

I just get confused when it's fractions would it be

(3/4)Ar^(3/2)?
 
Physics news on Phys.org
  • #2
You're close.. just the factor infront is incorrect. From your example, you know that the general rule is [tex] \int x^n dx = \frac{1}{n+1} x^{n+1} + C [/tex] Try and fit your specific case into this, and you should come out with the correct constant in front.
 
  • #3
defineNormal said:
Hello I'm trying to integrate Ar^(1/2)

would the answer be?

It depends what you are integrating with respect to. Are you integrating with respect to r?

well I know if X^2+2 were integrated it would be 1/3x^3

You forgot to integrate half of that function, your answer would be correct for integrating just x^2 but you didn't integrate the 2.

I just get confused when it's fractions would it be

(3/4)Ar^(3/2)?

If you are integrating with respect to r then this is almost correct, but your fraction isn't quite right. What is the rule when you integrate x to a power?
 
  • #4
well it's A*r^(1/2), for example a+bt+bt^2= a^2/2*bt^2/2*bt^3/3, I'm just not good w/ fractions, would it be Ar^(3/2)/(3/2)? or (3/2)Ar^(3/2)?
 
  • #5
defineNormal said:
well it's A*r^(1/2), for example a+bt+bt^2= a^2/2*bt^2/2*bt^3/3, I'm just not good w/ fractions, would it be Ar^(3/2)/(3/2)? or (3/2)Ar^(3/2)?

You're correct with Ar^(3/2)/(3/2) but you need to add the constant of integration, so the solution is (2/3)*Ar^(3/2) +C
 
  • #6
You are takinb calculus and you are telling us you cannot add [itex]1+ \frac{1}{2}[/itex]?? Am I misunderstanding this?
 

1. What is the process of integration?

The process of integration involves finding the antiderivative of a function. This is the reverse process of differentiation, where the derivative of a function is found. Integration is used to calculate the total change or accumulation of a quantity over a given interval.

2. How is the integral of a polynomial calculated?

The integral of a polynomial is calculated by applying the power rule, which states that the integral of x^n is (x^(n+1))/(n+1), where n is any real number except for -1. This rule is applied to each term of the polynomial, and the resulting terms are summed together.

3. What is the purpose of adding a constant when integrating?

When integrating a function, a constant term (C) is added at the end of the resulting function. This is because the derivative of a constant is always 0, so when finding the antiderivative, there could have been a constant term that was lost during the differentiation process. Adding the constant ensures that all possible solutions are accounted for.

4. Is there a difference between indefinite and definite integration?

Yes, there is a difference between indefinite and definite integration. Indefinite integration results in a function with a constant term, while definite integration results in a numerical value. Indefinite integration is used to find a general solution to an integral, while definite integration is used to find a specific value of an integral within a given interval.

5. How can integration be applied in real-world situations?

Integration has various real-world applications, such as calculating the area under a curve, finding the volume of a solid, and determining the displacement of an object over a given time interval. It is also used in fields such as physics, engineering, and economics to solve complex problems and make predictions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
445
  • Calculus and Beyond Homework Help
Replies
25
Views
351
  • Calculus and Beyond Homework Help
Replies
5
Views
971
  • Calculus and Beyond Homework Help
Replies
3
Views
277
  • Calculus and Beyond Homework Help
2
Replies
44
Views
4K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
547
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top