A different type of integral, if you want to call it an integral, is a "path integral". These are actually defined by a "normal" integral (such as a Riemann integral), but path integrals do not seek to find the area under a curve. I think of them as finding a weighted, total displacement along a curve.
The integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f (x)=C will have a slope of zero at point on the function.
Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.
it seems obvious that this integral is zero and so is the limit but what theorem we are using here? I see it's connected to Riemann sums with an interval=zero Right ? The function $\\mathrm{f}$ is
The noun phrase "improper integral" written as $$ \int_a^\infty f (x) \, dx $$ is well defined. If the appropriate limit exists, we attach the property "convergent" to that expression and use the same expression for the limit.
Wolfram Mathworld says that an indefinite integral is "also called an antiderivative". This MIT page says, "The more common name for the antiderivative is the indefinite integral." One is free to define terms as you like, but it looks like at least some (and possibly most) credible sources define them to be exactly the same thing.
A different approach, building up from first principles, without using cos or sin to get the identity, $$\arcsin (x) = \int\frac1 {\sqrt {1-x^2}}dx$$ where the integrals is from 0 to z. With the integration by parts given in previous answers, this gives the result. The distance around a unit circle traveled from the y axis for a distance on the x axis = $\arcsin (x)$. $$\arcsin (x) = \int\frac ...
One possible interpretation: a "normal" integral is simply a line integral where the path is straight and oriented along a particular axis. Thus, as soon as you perform a transformation to the integrand to make the path straight and oriented properly, you're back at a "regular" integral.
