Why is the area under a curve between two points equal to the difference of the antiderivative of the curve evaluated at the two points?One question I've never been able to satisfy myself with an answer to is why the area under a curve is equal to finding the curve's antiderivative, evaluating the antiderivative at two endpoints, and taking the difference of those two values (also known as taking the definite integral of the curve from point a to point b). It's a pretty basic calculus question, and even though my intuitive understanding of this question was never quite satisfied, it hasn't really hindered my mathematics education. I've just accepted that the area under the curve is as such. Well recently, I had a conversation with a friend, Philip Shao, and I brought up this question. He had an answer for me and patiently helped me gain the intuitive understanding I had always lacked. So prior to this conversation, here's what I understood:
Okay, so I understood the concepts above. If I'm just evaluating an integral (a problem such as "take the integral of this expression"), I could do that. That's just applying rules of integration, which can be derived from differentiation rules. And if I had to evaluate a definite integral, that made sense, too. I just found the expression for the antiderivative, plugged two values into the antiderivative, and took their difference. That's how a definite integral was defined, so I was just following rules. What I didn't understand was the geometric interpretation of a definite integral. I could apply the definition of a definite integral and evaluate a definite integral, but I didn't understand why this value was equal to the area under the curve between the two points a and b. You have a function F(x) which you're taking a definite integral of. If f(x) + c = the antiderivative of F(x), F(x) is the graph of the slope of f(x) at every point x. Okay, so since integration is the opposite of differentiation, if you differentiated f(x) to get F(x), then by integrating F(x), you get f(x). Okay, fine, this makes sense. Now you have two points a and b. Sure, I can evaluate f(b) and f(a) - that's just taking the value of the function f(x) at two points. No problem there. I can take their difference f(b) - f(a). No problem. It makes sense that this difference is the change in f(x) between a and b. But how is this difference also equal to the area under the curve of the derivative of this function f(x)? This was not intuitive or obvious to me, and this is what Philip successfully explained to me. After my conversation with Philip, this is the understanding I now have. For reference, let us refer to the diagram below:
In the following paragraphs, F(x) will mean the curve shown and f(x) will mean the antiderivative of F(x). So, say we want to find the area under the curve F(x) between 0 and x0. That's the shaded blue part in the picture above. My question was why this area is equal to the definite integral of F(x) from 0 to x0. Well, suppose there was an area function A(x), such that A(x) = the area under the curve of F(x) from -∞ to x. Then I'd agree, A(x0) - A(0) = area under the curve F(x) between 0 and x0. The definite integral of F(x) from 0 to x0 is equal to f(x0) - f(0), and supposedly, this is also equal to the area under curve F(x) between 0 and x0. If this is true, then A(x0) - A(0) = f(x0) - f(0), which would lead one to speculate that A(x) and f(x) are the same function. If I can show that they in fact are the same function, then I will have convinced myself that the area under the curve F(x) between 0 and x0 is equal to the definite integral of F(x) from 0 to x0, that is, the area under the curve = f(x0) - f(0). The way I argue that A(x) and f(x) are the same functions is as follows: The area under the curve F(x) between 0 and x0 is the shaded blue part above. Now suppose we want to add to this area the area under the curve F(x) between x0 and x0 + dx. Well, we can approximate this area with a rectangle (as shown) of width dx and height F(c), where c is some point in the interval [x0, x0 + dx]. The area of this rectangle is dx * F(c). Now, as dx --> 0, the rectangle becomes a better and better approximation of the area under the curve. Also, as dx --> 0, F(c) --> F(x0). The rate of change of the area function A(x) = (change in area) / (change in x) = dA/dx. Then, the instantaneous rate of change of the area function at x0 is the limit as dx --> 0 of dx * F(c) / dx = F(x0). From this, we can conclude that at every point, F(x) = the rate of change of the area function = derivative of the area function. Thus, the area function can be found by finding the antiderivative of F(x), which is precisely f(x). Thus, f(x) = area function A(x), and so the area under curve F(x) between a and b is precisely f(b) - f(a). Perhaps some of you reading this already had an intuitive understanding of what I explained above. But I didn't. And I had been wondering about this ever since I learned that the area under the curve between two points was the same as the definite integral evaluated between those two points. Finally, I have gained the intuition I was seeking. It's nothing profound, but for me, finally filling in this gap in my understanding was very satisfying. Thanks, Philip. P.S.: While making the figure above in PowerPoint, I utilized something I recently discovered. It used to bother me that when I was moving objects around, I couldn't place objects at exactly the place I wanted. I would click on an object and use the arrow keys to move it around, but it would only move in a certain-sized increment. Sometimes, though, I wanted movement in smaller-sized increments, but I thought PowerPoint restricted you to their increment size and you could not get any finer tuning. Recently, though, I discovered that if you hold down the 'Control' key while pressing the arrow keys, the object moves in smaller-sized increments. I also discovered that if you need even finer tuning than that provides, you can always right-click --> Size and Position, and tune the position of an object to the nearest hundredth of an inch. I feel kind of silly that I never realized before that such fine-tuning was always available. Yay math and PowerPoint. Two great things.
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First time selling something on Amazon (or online, for that matter)On Wednesday, February 3, 2010, I went to a Microsoft Tech Talk, at which the Microsoft recruiter raffled off lots of free stuff. I was fortunate enough to get a geeky T-shirt:
The T-shirt I wore a couple times and then I decided it was too small for me. It was really tight to begin with and the sleeves were a bit short - I finally checked the size label and it was actually a Ladies' Medium size. No wonder. I can get Microsoft Visual Studio 2008 Professional Edition for free through the Yale Software Library and the Microsoft Developer Network Academic Alliance program, so I decided to sell it online. After reading that Lily was successful in selling some Microsoft software she had on Amazon, I decided to do the same thing. At first, I got really excited when I saw Visual Studio 2008 Professional being listed on Google for over $1000. But then when I checked on Amazon, people were selling it new from about $600 and used from about $300. Most of the sellers were featured sellers with thousands of sales and > 95% positive feedback, so I realized it would be hard to compete with these sellers if I sold at around the same price. Thus, I chose a sale price of $555.55, which became the new lowest NEW price. I also made a little logo for myself and filled in some information on my storefront to make me seem more legitimate. Here is my logo:
I first listed my item during my spring break on March 13, 2010. However, even after three weeks of being the lowest NEW (as opposed to USED) seller, still no one bought it. At this point, I just wanted to get rid of it; considering I got it for free, anything would be good. So I made a drastic cut last night, lowering the price to $444.44, and almost immediately, someone bought it. After Amazon took its ~15% commission, my net earnings were $379.42. Score. Enough to fund a few visits to my friends at other colleges, eh? Don't want something? Sell it online!
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Ups and DownsMan, everything in life really is relative. Last week, I played clarinet in the pit orchestra for the musical Man of La Mancha, directed by my friend, fellow bandie, and fellow Stilesian Kelsey Sakimoto. We had three four-hour rehearsals each night - Monday, Tuesday, Wednesday - and then three shows - Thursday, Friday, Saturday. The play was really inspiring. To dream the impossible dream. That one man should be struck down ten thousand times, he must rise again and do battle. It's such a moving play, with an idealistic message, and it really appealed to the dreamer in me. All week, I had the songs from the musical stuck in my head. I watched so many YouTube clips of theater and high school productions of Man of La Mancha. After rehearsals, I wasn't able to do any work, partly because it was late and I was tired, but mainly because I was still wrapped up in the musical. Really, it's such a poignant musical and the idealist in me fell in love with the "lunatic" characters. Now, the show is over and it's Sunday night and the last week of classes is about to begin and I have work to do. Sigh, back to the normal life. This is the normal life, but it just seems so depressingly dull after my spirits were buoyed up so high by Man of La Mancha. I think this must be what it's like to take drugs or stimulants. You get really high and feel so good and then the effect wears off and you crash back to earth really hard and feel like shit. Everything really is relative. After experiencing such intense joyful emotion, going back to the regular routine doesn't feel normal, it actually feels dull and boring. I am reminded of something Andy Tien said to me in a GTalk chat recently: "Real life is so up and down. I learned to hold on though and enjoy the ride. Maybe loosen my hold a little and get some air from time to time though. There's nothing like flying, however short it is." I feel like last week, rehearsing and playing in the musical was like my moment of flight. But now I've come down and landed and am trodding again on foot. It's really been an up and down. Sigh. Just gotta pull through these last few weeks and then it's summer :). Further commenting has been disabled due to spam. |