Indefinite Integral of a contant function

  • Thread starter rambo5330
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  • #1
rambo5330
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This question is bugging me

if an indefinite integral is of the form F(x) + C
stating that F(x) is an antiderivative of f(x) and since the derivative of a constant is 0 the collection of all of the anti derivatives are of the F(x) + C accounting for the fact any constant can be tagged to the end of that anti derivative..
now if this were an initial value problem you can take F(0) = 10 and solve for C correct...

well why with a constant function say f(x) = 1 when I solve for indefinite integral of the form F(x) + C .... I get x + C
well if f(x) is a constant function then wouldnt f(0) = 1 f(5) = 1
but in each of these cases i get C + 5 = 1 therefore c = -4 or C + 0 = 1 therefor C = 1

but if this if F(x) + C is a general expression that allows you to find the area under a graph up to point x ... then
find the area under a function which is constant f(x) = 1 on interval [0,9]
should = 9

but x + 1 = 9 + 1 = 10 so its obvious to see that C should equal zero

where am I going wrong can someone please clear this up?
 

Answers and Replies

  • #2
Clever-Name
380
1
You seem to be getting confused as to what F(x) + C and f(x) are. f(x) is the function you're integrating, so yes C can be any constant in the end. It can even be 24937203. It doesn't matter. If you're trying to find the area of the function f(x) = 1 from 0 to 9 then you are no longer solving an INDEFINITE integral, now you need to set limits on your integrate and solve the DEFINITE integral.

For this case I'll just show you the result:

[itex] \int_{0}^{9}1dx = x + C|_{0}^{9} = (9 + C) - (0 + C) = 9 + C - C [/itex]

So you see the C's will cancel out in the end, it doesn't matter what they are because when solving the definite integral that constant will ALWAYS end up cancelling out.
 

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