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What does it mean for an integral to be convergent? The improper integral $\int_a^\infty f(x) \, dx$ is called convergent if the corresponding limit exists and divergent if the limit does not exist While I can understand this intuitively, I have an issue with saying that the mathematical object we defined as improper integrals is "convergent" or "divergent"
Calculus proof for the area of a circle $\begingroup$ @andreas vitikan Note that your teacher's approach is what Jennifer Dylan describes above, and while the calculation of that integral is more difficult, the idea behind it is quite straightforward On the other hand, the integral you have is quite easy to calculate but the background is not as intuitive, IMHO
integration - Improper integral of sin(x) x from zero to infinity . . . I was having trouble with the following integral: $\int_{0}^\infty \frac{\sin(x)}{x}dx$ My question is, how does one go about evaluating this, since its existence seems fairly intuitive, while its solution, at least to me, does not seem particularly obvious
Integral of $\\sqrt{1-x^2}$ using integration by parts This integral is much interesting since it can be interpreted in three ways as far as I know Mentioned by the OP, trigonometric substitution can be applied Integration by part Geometric interpretation mentioned in the comment which I think is the most intuitive I will give each a reasonable explanation in the following Trigonometric
calculus - Is there really no way to integrate $e^{-x^2 . . . $\begingroup$ @user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral
Can a limit of an integral be moved inside the integral? The main result gives a necessary and sufficient condition under which the limit can be moved inside the integral The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability
Integral of Complex Numbers - Mathematics Stack Exchange $\begingroup$ If I set limits on the integral then that would refer to area under the curve but why does it come out to be different in the real and complex scenario $\endgroup$ – Sajid Rizvi Commented Jul 28, 2017 at 20:15