I recently received a message from a friend and the heading of the post perfectly describes what was said to me. Thereafter, an interesting integration question was sent to me. It read as follows:

I must admit, it does look quiet scary. My immediate thought was that some sort of substitution was required but I really had no idea as to where and how this should be done. Two pages and a headache later, I thought to myself why don’t I get a rough idea of what the answer should look like. Once again, let’s start approximating things (as it turns out, my approximation gets me the absolutely correct answer).

I looked at the integration bounds and noted that the point x = 3 was in fact the midpoint. I then decided to construct a Taylor polynomial for the above function around the point x = 3. It was done as follows:

You might be wondering why I didn’t bother taking anymore derivatives. It’s because I wanted something simple to work with and it’s quite clear that the derivative of the above function will be tedious to calculate and most probably result in an irrational value at the point x = 3. So let’s simply make this statement:

Now let’s take the integral of the first term of the series and see what value we arrive it.

A value of 1 seems pretty exact, doesn’t it? Let’s try and evaluate the actual integral in order to see just how close my approximation is.

Let

And now we need to prove a result that I find to be rather useful. We will do this as follows:

Let

So what does the above proof tell us? It says that for our original function, the following is true:

We can now add both of the integrals because they have a common denominator. Therefore:

And now,

Wow! It seems like my approximation was absolutely spot on! I can think of a reason as to why this may have happened. The single term of a could have been under approximating the function to the left of the point x =3 and the opposite could have happened to the right of the point x =3; this could have resulted in any inaccuracy being eliminated over the interval [2, 4]. This would then mean that the mean value over the interval would have been a . Why don’t we check this quickly because I’m really intrigued by this result.

The mean value would be:

My assumptions were in fact correct! I find it quite remarkable that one question managed to make me think about a few different aspects of Mathematics. Additionally, it seems like I’ve stumbled across a result that I haven’t really thought about. The result is that the integral of a function over an interval is equal to the integral of the mean value of that function over the same interval. Let me know what you think about this result in the comments below. Lastly, I hope everyone enjoyed reading my review of the question as much as I enjoyed tackling it.

MohammedDecember 30, 2016 at 2:19 amI taught you well my boy❤