Indefinite Integrals

  • #1
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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
 

Answers and Replies

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

Daniel.
 
  • #3
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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
dextercioby
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Explain to me the first formula.Why doesn't it include "a" & "b"?

Daniel.
 
  • #5
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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
jtbell
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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
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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
dextercioby
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Do you know the Fundamental Theorem of Calculus...??

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

Daniel.
 
  • #9
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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
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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
dextercioby
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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
dextercioby
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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
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[tex] C = -2u^2 [/tex] Is this right?
 
  • #14
dextercioby
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What has "u" gotta do with your integral,which is done wrt "x"?Better said,who is "u"??

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

Daniel.
 

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