This is a post from my old blog, “Ultramega”. I’m squeezed on time, but I wanted to post something. From time to time I may repost some of the stuff I wrote there last year. This, for example, is from last year when I was neck deep in Calculus…
[So] I was working on was finding the values of c guaranteed by the Mean Value Theorem(MVT) for Integrals for the function f(x) = x - 2√(x) over the interval [0, 2].
Since I know everyone was insatiably curious about this, here it is.
For starters, in order to apply the MVT, the function f(x) = x - 2√(x) needs to be continuous over the interval, and it needs to be differentiable over the open interval, which in this case is (0, 2).
Well, as x approaches 0, f(x) approaches 0 also, and then increases, along with x, to infinity. So it’s definitely continuous on [0, 2].
Also, it is differentiable. For a function to be differentiable at a point, we are basically simply saying that the function has a specific slope at that point. If the function is differentiable over an open interval, then we are saying that on every point on that interval, the function has a specific slope. Hopefully it’s clear that this function is also differentiable on this interval, because it is.
Now what does the Mean Value Theorem say? In a nutshell, it wants us to believe that if the function fulfills the two conditions (above), then there exists a number (or numbers) c such that ∫ab f(x)dx = f(c)(b - a). (Note that ∫ab does not in this case represent the integral of a raised to the power of b, but denotes the limits of the definite integral, a to b — HTML sucks for displaying math problems).
To actually find c, we multiply both sides of ∫abf(x)dx = f(c)(b - a) by 1/(b - a), so we arrive at [1/(b - a)]∫ab f(x)dx = f(c).
Now we can get our hands dirty. Plugging in 0 for a and 2 for b (we get these from the interval given right at the very beginning, [0, 2]), we get:
[1/(2 - 0)]∫02 x - 2√(x)dx = (1/2)∫02 x - 2√(x)dx
Integrating this, we get,
(1/2) [ (x2/2) - (4/3)x3/2 ]
evaluated from 0 to 2, or
(1/2) [ (22/2) - (4/3)23/2 - 0 ]
which we can simplify to
1 - [ ( 4√(2) )/3) ]
Whew! Now the above is really equal to f(c), according to the Mean Value Theorem, which means it must be equal to c - 2√(c) (see the original function at the top). So,
1 - [ ( 4√(2) )/3) ] = c - 2√(c)c - 2√(c) = [ 3 - 4√(2) ]/3
To make this easier, let’s say that √(c) = z. Now we have
z2 - 2z = [ 3 - 4√(2) ]/3
Why, the left side is a quadratic equation! We can add one (to each side) to complete the square and get
z2 - 2z + 1 = [ 3 - 4√(2) ]/3 + (3/3)z - 1 = -+ √[ ( 6 - 4√(2) )/3 ]
z = 1 -+ √[ ( 6 - 4√(2) )/3 ]
But z = √(c), so we can now back-substitute and get
√(c) = 1 -+ √[ ( 6 - 4√(2) )/3 ]c = ( 1 -+ √[ ( 6 - 4√(2) )/3 ] )2
which we can finally just plug into our calculators to get 0.4380 and 1.7908.
Yes, it’s true. I just posted a calculus solution, with explanation, to my weblog. I’d say it was a big waste of time, but believe it or not, explaining the problem seems to have helped me understand it more.
Maybe I need to start a mathblog…
