Integral using Substitution
Date: 4/11/96 at 4:3:39 From: Anonymous Subject: integration by substitution Dear Dr Math, I am a year 12 student at high school in Sydney, Australia and I have a math exam next week. I was wondering if you could help me with this problem. Using the given substitution evaluate the following correct to 3 decimal places. xdx -------- 2 1/2 ( x + 1 ) 2 2 given u = x + 1 I don't know if you will be able to understand my question because I don't have the right symbols. 2 u this means u squared and the question is supposed to be preceded by an integral sign with the limits of 3 and 0. The question in written form is: The integral of xdx over x squared plus one all square rooted. I hope you can help. Also do you know of any math pages located in Australia? Thanks a lot! Bye, Tracey Pearce
Date: 4/12/96 at 23:47:40 From: Doctor Syd Subject: Re: integration by substitution Dear Tracey, Your question is very clear, don't worry! Ususally we notate taking x to the second power by this: x^2, and that is the notation I'll use since it is much shorter. Okay, so to solve the problem, you let u = x^2 + 1. Everything inside the integral is in terms of x or dx, right? So, all we need to figure out before we begin the actual substitution is what dx is. If we differentiate both sides of the equation u = x^2 + 1, we get du = 2xdx, right? So, using these two equations, let's make the substitution: xdx = du/2 since du = 2xdx (x^2 +1)^1/2 = u^1/2, So we don't have to deal with a fraction, let's write what is in the integral as: (xdx)(x^2 + 1)^-1/2 Then, what is inside the integral ( I think it is called the integrand if I am remembering correctly!) is: (du/2)(u ^-1/2) This is something that is much easier to integrate, right! When evaluating the integral, you must be careful, since your limits are in terms of x, but your integral is in terms of u. You have two options: 1) As soon as you subsitute the u's for the x's, make similar substitutions in your limits. So, in your problem, since the limits of the integral are x=0 and x=3, plug these values into the equation relating x and u to get the corresponding values for u, and use these new values as your limits in your new integral. Or, 2) Don't worry about changing your limits. Proceed as you normally would when integrating something, except when you are at the stage where you are ready to plug in limit numbers to what you have found to be the antiderivative of the function you are integrating, STOP! At this stage, convert your antiderivative that is in terms of u's back into something that is in terms of x's using the original formula you used to convert from x's to u's. Then, go ahead and evaluate using your old limits. So, I think the rest of the problem is probably pretty doable. Follow one of the above two methods to get your answer! You probably learned the chain rule for calculating derivatives. What you just did in this problem was use an anti-chain rule of sorts. Have you talked about the anti-chain rule in class? It is a helpful tool, so think about what an anti-chain rule might say, and how you might prove it. If you want to know more about this, feel free to write us back! Hope this was helpful! -Doctor Sydney, The Math Forum
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