I was wondering,how do we calculate the perimeter of a region using integral calculus?I know that to calculate the area we have to draw the region and if we...
can someone give me some really hard intergrals to solve? make sure they are in the range of calculus 1-2 (anything before multivariable) My teacher assigned some few hard integrals, and they are fun. I want to try moer. thanks.
I want to find the solution to the integral ##\theta = \int_0^ {\theta}\frac {du} {\sqrt { (c-u^2 +2u^3)}}## I can see that ##\frac {d^2u} {d\theta^2} = A...
In my paper on renormalisation I mentioned what most who have studied calculations in Quantum Field Theory find, its rather complicated and mind numbing.
Splitting improper integrals is often necessary to handle discontinuities and infinite bounds effectively. While it may seem tedious, dividing the integral at points of discontinuity ensures the existence of limits and prevents divergence. The discussion highlights that convenient split points are typically at the origin or at discontinuities, allowing for clearer limit evaluations. Using ...
Evaluating a fourth-order integral, such as f(x,y,z,t) dx dy dz dt, generally represents a hypervolume in four-dimensional space, extending the concepts of area and volume from double and triple integrals. While it can be interpreted physically as space-time, this interpretation depends on the context of the function being integrated. The complexity of integrating in a space-time framework ...
Since I'm trying to make sense of the application of a concept, I figured I'd post this thread here. If this belongs in the homework forum, I sincerely apologize. For the longest time I would do integral problems in Calculus. I figured I understood how to do them because after I finished...
Homework Statement I know that a single integral can be used to find the area under a y = f(x) curve, but above the x axis. Correct me if this example of a double integral is invalid: If I hold a piece of paper in mid air and it droops, the double integral will give me the volume of the object...
Why is it that when we evaluate a surface integral of: f(x, y ,z) over dS, where x = x(u, v) y = y(u, v) z = z(u, v) dS is equal to ||ru X rv|| dA Why don't we use the jacobian here when we change coordinate systems?
I have a question about work integrals. I'm trying to reconcile using integrals to essentially multiply force by distance, but the fact that there appear to be multiple different types of problems that seem to be fundamentally different is making it difficult. Here are some example problems...
