Line Integrals in Wolfram Alpha?

Sure does. Here’s how you use it.
Calculate the area under a curve from 0 to some point by entering that value into Wolfram Alpha’s calculator and changing the input parameters: Area (sqrt(n)) Enter pi ‘= 3’ Then hit “digits” until reaching 1 over 10 or higher if needed..

Area (sqrt(n)) Enter pi ‘= 3’ Then hit “digits” until reaching 1 over 10 or higher if needed.

Bingo! If you get a result of 0, it’s simply telling you that the number entered is less than one. This is useful for showing your teacher that there is no definite answer to a question on an exam.

Wolfram is a powerful tool that many people don’t know exists, so go try it out and let us know how you’ll use the app to your advantage.

Wolfram Alpha can calculate a line integral, but the form that has to be used is not specified.

When I was in college, my favorite course to take for credit was Complex Analysis. It’s not often that you come across an integral which is both intuitive and interesting! In this case though it seems as if the author has done just what he set out do – make something complicated seem simple enough so that even those without any knowledge of complex numbers can understand its purpose or value behind using these types mathematical equations on real valued functions.[1] says that we can think of the integral as a function on our complex plane, and then use some simple math to find out where it is integrating. The first step in understanding this process involves considering what happens when you treat an integrand as if they were functions within two dimensional space; for instance cos z/[(z+1)^2 + 2](z+) = Re f(z).The first part of this integral is similar to what you would do in high school. You take a semicircular arc with nonzero imaginary number tracing between (0, Pi) and R at 0 when parameterized by Cos t or Sin T respectively which are on opposite sides but close together so they kind-of cancel each other out; then there’s also an overreal line segment that has nothing special about it – just think back how many times we had Integrals Over Real Segments.

Infinity is the limit of this integral. We can show that because e*(iz) has a finite upper bound, our circular part must go to zero as R-infinity . As such when computing values inside those limits it’s only relevant what happens on one side – in other words ,the real axis will always yield some value during calculation due solely from its proximity with regards towards infinity.

The key to my solution is the concept of a ‘small’ circle, this type of analysis is provided by Mihalis Dafermos , whose paper on limit representations of analytic functions  I used as a direct framework for developing a simple solution. I recommend giving his article a read before going further with mine… but anyways let’s get to the meat.

Here’s a graph of the function we’re going to begin by computing:

This graph is color-coded in a way that distinguishes between components of the complex plane, in fact you can think of this graph as a piece of virtual paper with coordinates in it. The real axis being red, and the imaginary axis having a blue color to it. Each point in this virtual coordinate paper will produce a real valued result after the calculus routine I sketch out here is complete- all that’s needed is to interpret it into geometric terms, and then you’ll see what’s going on.

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