Question

The definite integral

Why do we not take into account the constant of integration when doing questions with a definite integral?

Answer 1

Such as area under a curve

Answer 2

read this link: http://math.stackexchange.com/questions/156376/definite-integral-and-constant-of-integration

Answer 3

I.e Area under graphs

Answer 4

you will find that in trying to evaluate a definite integral you will subtract the constant.

Answer 5

Right. . So isn't it good to write it ... or I guess it must be like a x and 1x thing

Answer 6

i find this link confusing [doesn't mean it is, of course]
\[\int_{0}^{x} \ f(t) \ dt = F(x) - F(0) = F(x) + C\] if defined at 0

how does that help illustrate anything?

i refer specifically to this bit:
"In the case above, ∫x0f(x)dx, there is confusion because the same variable is used inside the integration as in the bounds. The bound variable x inside the integral is not the same as the free variable x in the limit. To reduce the confusion, your integral can also be written as ∫x0f(t)dt by renaming the bound variable. In any case, this is a definite integral."

Answer 7

to wit

https://gyazo.com/98df70711ab8ef2676e382dcbf9cf2ce

Answer 8

it depends on the reason for integrating.

if you are trying to find an antiderivative, then there are a family of functions that will work, and they differ by a constant.

if you are trying to determine the area between curves ... you already have functions defined for you and you are actually working a limiting process of say a Reimann sum.

Answer 9

what set of functions have a derivative of: 2x?

f(x) = x^2 + C
-----------------------

what is the area from x=0 to x=2, between the curves: y=2x and y=0?

(2)^2 - (0)^2 = 4

Answer 10

im not sure if the definite integral can be considered as some initial value related thing. but i think it has more to do with the way it works out from the limiting definition.

Answer 11

hmm, try a test ... use the +C

\[\int_{0}^{2}2x~dx=\large [x^2+C]_{0}^{2}=2^2-0^2+(C-C)\]
maybe, but i dont think its rigorous.