Wednesday, September 18, 2013

Solution to the cool integration by parts problem

This problem requires a bit more ingenuity.
Start out using integration by parts as you normally would.
You may notice that the new integral acquired still poses the same issue we started out (but with cosine instead of sine). Let us do integration by parts once more and see what happens
Now I need to substitute this back in. I need to also make sure to distribute the negative.
Now we have the exact same integral we started out with on both side. all we really need to do is add the integral of e^(x)sin(x) to both sides of the equation. Then we'll have 2 times the integral of e^(x)sin(x) on one side. Simply divide by two and the whole thing will be solved.
Is your mind blown or what?




Tuesday, September 17, 2013

A cool integration by parts problem

Here's a cool integral that requires integration by parts. It's a little tricky, but the solution is mind blowing.


Thursday, September 12, 2013

Deriving the volume of a sphere solution

note* I'm assuming you are already familiar with calculus. If you're not, it's probably not going to make any bit of sense.

I will be doing this using integration techniques. Remember learning about solids of revolution from calc. I? That's what I'll be using. I'm not really sure if it's the best way, but it works.

Consider the graph of a basic semi-circle. If you were to revolve it around the x-axis, the solid generated would be a sphere whose cross sections are circles.
Imagine slicing through a sphere. You'll notice that the cross sections are circles of varying sizes. What I'll essentially be doing is creating a summation of disks that each have a very small thickness of dx.
To start off, I'm going to create a function of the cross sectional area and integrate it from one end of the semi circle (-r) to the other (r). I'll need the formula for the area of a circle. The radius of each cross section will vary as a function of x, and so I need to make the appropriate substitution for the radius. Through integration, the disks of varying cross sectional area will be summed together to create a sphere.
There's nothing all that tricky about solving this integral. If you work it all out you should end up with the correct formula for the volume of a sphere.




Wednesday, September 11, 2013

Deriving the volume of a sphere

For this next one I want to derive the volume of a sphere using calculus. I don't have much more to add, but I feel like I should at least post a picture of something...

Tuesday, July 30, 2013

Sunday, July 28, 2013

problem 5: solution to the slightly harder integral

I must say, this was a bit more challenging, but nothing I can't do.


Using properties of logarithms, I can rewrite the integral like this:
It simplifies to this:
Now I just need a little u-substitution. 
For u-substitution to work, I'll also have to rewrite the integral as follows:
After doing u-substitution, it should look like this:

The integral of ln(x) requires integration by parts. I'm not going to go through the process now because I did it in a different post. It should like this this:
Well, it's simple from here. Just plug back in (x^2+1) in place of u. I'm going to leave it in terms of u because it's long and messy in terms of x.

Thursday, July 25, 2013

problem 5: A slightly harder integral

My brother said this problem was on one of his Calc. II exams and challenged me to do it. Challenge accepted