Definite integrals

  • Thread starter PFuser1232
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  • #1
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I have a few questions about the following property of definite integrals:
$$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$$
What exactly are the prerequisites for this property?
Should ##c## be a member of ##[a,b]##? Should the function ##f## be defined at ##c##?
 
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  • #2
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c need not be within [a,b]. Itis because c is going to be subtracted
 
  • #3
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c need not be within [a,b]. Itis because c is going to be subtracted
Could you please elaborate? And what about the existence of ##f(c)##?
 
  • #4
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If f'(x) is differentiation of f(x) with respect to x, then

∫ f'(x) = f(x) + c

And within limit b to a it will be f(b)-f(a)

Look at the right hand side, there is a sum of two integrals having limit c to a, and b to c.
So R.H.S={f(c)-f(a)} + {f(b)-f(c)}

So the f(c) is subtracted and finally it is f(b)-f(a) like the L.H.S

So, c need not be within [a,b]

I am not sure if f(c) should be defined.
As f(c) is getting subtracted so it does not matter. But generally an equation is used for solving problems. If you take such value of c where the function is not defined, you cannot use the equation to solve the problem.
 
  • #5
HallsofIvy
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c must be such that f is defined and differentiable on some interval containing a, b and c. For example, if a< b< c, then we have [itex]\int_a^b f(x)dx+ \int_b^c f(x)dx= \int_a^c f(x)dx[/itex] so that [itex]\int_a^b f(x) dx= \int_a^c f(x)dx- \int_b^c f(x)dx= \int_a^c f(x)dx+ \int_c^b f(x)dx[/itex]. I have no idea what firefly meant by "c is going to be subtracted'.
 
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