How Do You Set Up Indefinite Integrals for Different Functions?

In summary, the first two formulas are correct and the third is incorrect because the derivative of the function placed inside the integral, and the original function is on the right side of the equal sign.
  • #1
courtrigrad
1,236
2
Hello all

If you have the funtions: [tex] y = x^2, y = sin x, y = 2x^4 [/tex] and you are asked to set up the corresponding integrations, would it be:

[tex] \int^b_a 2u du = x^3 [/tex]
[tex] \int^b_a -\cos u du = \sin x [/tex]
[tex] \int^b_a 8u^3 du = 2x^4 [/tex]

Thanks
 
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  • #2
Explain why you think your 3 formulas are correct.Personally,i find them wrong...

Daniel.
 
  • #3
I think they are correct because the derivative of the function placed inside the integral, and the original function is on the right side of the equal sign.

Isnt't it true that the indefinite integral [tex] \Phi (x) = \int^x_a f(u) du [/tex] of a continuous function [tex] f(x) [/tex] always possesses a derivative [tex] \Phi'(x) [/tex] and [tex] \Phi' (x) = f(x)[/tex]?

Thanks
 
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  • #4
Explain to me the first formula.Why doesn't it include "a" & "b"?

Daniel.
 
  • #5
Oh I see

This is an indefinite integral so it should be:

[tex] \int^x_a 2u \ du = x^2 [/tex]

Is this right?
 
  • #6
courtrigrad said:
Oh I see

This is an indefinite integral

Indefinite integrals don't have limits. Definite integrals are the ones that have limits. What you have is a definite integral.

so it should be:

[tex] \int^x_a 2u \ du = x^2 [/tex]

Is this right?

No, you still haven't taken account of the lower limit, a, of your integral.
 
  • #7
ok so [tex] \int^x_a 2u \ du = x^2 [/tex] or [tex] \int^b_x 2u \ du [/tex] and we ahve [tex] \int^x_a 2u \ du +C = x^2 [/tex] [tex] \int^b_x 2u \ du +C [/tex] . Is this correct?
 
  • #8
Do you know the Fundamental Theorem of Calculus...??

What u have written is total nonsense...Definite integrals do not have integration constants...

Daniel.
 
  • #9
What are you talking about?? Isn't [tex] F(x) = c + \int^x_a 2u \ du = \int 2x \ dx [/tex] an indefinite integral? Let's say we know the the indefinite integral is [tex] \Phi (x) = \int^x_a f(u) \ du [/tex] and we know a primitive function [tex] F(x) = \int f(x) \ dx [/tex]. Then [tex] \Phy (x) = F(x) + c [/tex]. But [tex] \int ^a_a f(u) \ du = 0 [/tex] and [tex] 0 = \Phi (a) = F(a) + c [/tex] and solving for c we get [tex] c = -F(a) [/tex] and [tex] \Phi (x) = F(x) - F(a) [/tex] So to evaluate the definite integral [tex] \int^b_a f(u) \ du = F(b) - F(a) [/tex]
 
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  • #10
I think I got it: IT would simply be : [tex] \int 2x \ dx = x^2 [/tex] or [tex] \int 2x \ dx = -x^2 [/tex] because they differ only by an arbitrary constant.
 
  • #11
No.A definite integral is
[tex] \int_{a}^{b} f(x) \ dx [/tex]

and an indefinite one is
[tex] \int f(x) \ dx [/tex]

Daniel.
 
  • #12
courtrigrad said:
I think I got it: IT would simply be : [tex] \int 2x \ dx = x^2 [/tex] or [tex] \int 2x \ dx = -x^2 [/tex] because they differ only by an arbitrary constant.

No,what u've written is wrong...

[tex] \int 2x \ dx =x^{2}+C [/tex]

Find the constant C as to obtain your second formula...

Daniel.
 
  • #13
[tex] C = -2u^2 [/tex] Is this right?
 
  • #14
What has "u" got to do with your integral,which is done wrt "x"?Better said,who is "u"??

Daniel.
 
  • #15
oh ok so it would be [tex] C = -2x^2 [/tex] but this isn't a constant is it?
 
  • #16
Obviously no.Good thing that u eventually realized it by your own.

Daniel.
 

What is an indefinite integral?

An indefinite integral, also known as an antiderivative, is a mathematical concept that represents the reverse process of taking a derivative. It is the set of all possible functions whose derivative is equal to a given function.

How is an indefinite integral represented?

An indefinite integral is usually represented by the symbol "∫" followed by the function to be integrated, and then the variable of integration. For example, ∫x^2 dx represents the indefinite integral of the function x^2 with respect to the variable x.

What is the purpose of finding an indefinite integral?

The main purpose of finding an indefinite integral is to evaluate the original function from its derivative. It is also used in various mathematical applications, such as finding the area under a curve and solving differential equations.

What are some common techniques for evaluating indefinite integrals?

Some common techniques for evaluating indefinite integrals include substitution, integration by parts, and using trigonometric identities. It is important to choose the appropriate technique based on the form of the function being integrated.

How do indefinite integrals relate to definite integrals?

Definite integrals are used to find the exact numerical value of the area under a curve, while indefinite integrals give a set of possible functions that would result in that definite integral. In other words, definite integrals are a specific case of indefinite integrals.

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