Integrals and Trig Functions
Date: 4/6/96 at 16:2:52 From: Anonymous Subject: Calculus: integral from a to b of trig function Hey there! This is a problem for Advanced Placement Calculus, and if I solve it, I'll get extra credit. Evaluate EXACTLY: the integral from pi to zero of the square root of (1-sinX). I don't even know where to start and what should I use to solve this problem, so I would apreciate it if you would be able to give me some hints and starting points. That'll be really great. Thanks. Bye, Ksenya. 8)
Date: 4/7/96 at 19:49:1 From: Doctor Aaron Subject: Re: Calculus: integral from a to b of trig function Hi, Ksenya, It's hard to imagine how someone in a pre-calc class would do an integral and understand what they were doing. I'll give you some things to think about, and I hope they'll be helpful. First I'll define an integral. Let f(x) be a function that takes the domain (a,b) to the range (c,d). Let a<x1<x2<.....<x(n-1)<b. Now we're going to let x0 = a and xn = b. Here the 1,2,n-1,n should be subscript and denote indices. We could just use different variable names, but we're going to want to take a limit as n goes to infinity so unless we use indices, we'll run out of variable names. Then we can think about the partition of (a,b) as all of the subintervals: (x0,x1), (x1,x2), ... (x(n-2),x(n-1)), (x(n-1),xn). Then we can call an arbitrary interval (xi,x(i+1)). We can think of the group of subintervals as this arbitrary one where i ranges from 0 to n-1. The next thing is to pick a random point xij that is inside of (xi,x(i+1)). Now we get back to the function. We're going to look at f(xij). We multiply f(xij) by (x(i+1) - xi). Now we're ready for the definition of the integral: The _Integral_ of f from a to b is equal to the sum over all i of f(xij) times (x(i+1) -xi). You may be thinking that this is completely unintelligible and even if it is, why would anyone want to know about that awful sum. Well it might help to think of this geomoetrically. Get a piece of graph paper and draw the graph of your favorite function. Now pick an interval on the x-axis. An interval of length 5 with endpoints that are integers is good. Draw vertical lines at the endpoints of this interval. Only look at the piece of the graph that is contained within these vertical lines. Now pick a bunch of points inside of your interval on the x-axis. It's much easier if you pick integers. Now draw vertical lines at these points. Your graph should be partitioned into lots of little pieces. Pick a random point in each of these pieces. Now multiply the function value at that point by the length of the interval which contains it. Now the last step is to add all of these guys up. We call this a Riemann sum. Geometric Interpretation: We remember that the product of 2 numbers is the area of the rectangle whose sides have length equal to the numbers. Now draw a vertical line down from the function value at the random point. We're multiplying by the subinterval length, so in each subinterval on the graph of the function, we can draw a rectangle whose base is the subinterval on the x-axis, and whose height is the value of the function at the random point. If we do this in each subinterval, we get something that approximates the area under the graph of the function. An integral is defined to be the limit as n goes to infinity of this Riemann sum. Then we can say that the integral of a function from a to b is exactly the area under the graph of f from a to b. Your function is sqrt(1-Sin[x]). It sure would be a pain to have to do this whole process again, especially for such an annoying function. There is something called the fundamental theorem of the Calculus that gives a shortcut because life is too short to do Riemann sums, especially if we have to take the limit as n goes to infinity. The proof is hard, and even the statement of the theorem requires a knowledge of derivatives, but here it is: **Caution: this is a deep mathematical statement so don't use it without thinking about what it means.** If f is the derivative of F, then the integral of f from a to b is equal to F(b) - F(a). I think that if you think about the Riemann sum a lot, you will learn a lot of math, and even if you can't evaluate the integral exactly I'm sure your teacher will give you partial credit (no promises - I don't know him or her). Good luck. If you understand what's going on by your 5th reading you're doing great. -Doctor Aaron, The Math Forum
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