Why Does Integration by Substitution Work?Date: 12/13/2006 at 22:03:56 From: Reuben Subject: Why integration by substitution works I just learned the process of integration by substitution. My textbook seems to explain it for the most part, but it says "imagine that you multiply du/dx by the dx in the derivative to get du." This seems to work, but I've found online that technically, you can't treat differentials as numbers that can be "canceled out." A) Is this true? It seems to me that they're just very very very small numbers. So why could you not cancel them out? B) If that's the case, how do you prove that substitution works and use integration by substitution WITHOUT treating differentials as numbers? Date: 12/14/2006 at 08:39:02 From: Doctor Fenton Subject: Re: Why integration by substitution works Hi Reuben, Thanks for writing to Dr. Math. There are several ways to look at differentials, but probably the simplest is to view them as a formal bookkeeping device for keeping track of the constants when finding antiderivatives. Many of the integration (or antidifferentiation) rules are actually counterparts of corresponding differentiation rules, and this is true of the substitution theorem, which is the integral version of the Chain Rule. The Chain Rule says that if we have a composite function F(g(x)), and if f(x) = F'(x) is the derivative of the outer function F(x), then [F(g(x))]' = F'(g(x))*g'(x) = f(g(x))*g'(x) . The antiderivative version of this says that if we want to find the antiderivative of f(g(x)) and we know the antiderivative F of f, then the antiderivative of f(g(x)) is just F(g(x)), and we have reduced the problem of finding the antiderivative of the complicated expression f(g(x)) to that of finding the antiderivative of f, which we usually write with a different independent variable such as u. That is, the Chain Rule says that [F(g(x))]' = f(g(x))*g'(x) , (remember that F' = f) so / | f(g(x))*g'(x)dx = F(g(x)) + C . / But since F' = f, / | f(u)du = F(u) / (the letter used for the variable of integration is a dummy variable, so we can use any letter we wish). This lets us write / / | f(g(x))*g'(x)dx = | f(u)du / / where we understand the right side to be F(u) with u replaced by the formula g(x). So it appears that in the integral on the left side, we replaced g(x) by u and g'(x)dx by du (similar to the way it "appears" that the sun rises and sets). This process doesn't have to be considered meaningful in itself, but rather just a mnemonic or aid in determining the outer function f(x) in the composition. The differential part of this substitution can also be thought of as a mnemonic, since we can mix Leibnitz and function notation to write, if u = g(x), that du -- = g'(x) , dx and "multiplying" by dx gives du = g'(x)dx . One can make this procedure logically rigorous by introducing the concept of differential forms, but that requires a lot of mathematical machinery, so that it is easier to just think of it as a formal computation. (There are books such as Bressoud's _Second Year Calculus_, Edwards' _Advanced Calculus: A Differential Forms Approach_, and Spivak's _Calculus on Manifolds_ which go over this in detail.) For example, to integrate / | [x^3+1]^(1/2) x^2 dx , / we can notice that the integrand has the form of the derivative of a composite function f(g(x))*g'(x), with f(x) = x^(1/2) and g(x) = x^3+1, but g'(x) = 3x^2, not just x^2, so the constant is not quite right. However, if we multiply and divide by 3, we have / / | [x^3+1]^(1/2) x^2 dx = (1/3)| [x^3+1]^(1/2)*(3x^2)dx / / and the integrand on the right is exactly a derivative now, of the composite function (2/3)[x^3+1]^(3/2) (we can see this because the outer function f(x) is x^(1/2), whose antiderivative F(x) is (2/3)x^(3/2), so the antiderivative is (1/3)F(g(x))). This argument has computed the integral or antiderivative without substitution. To use substitution, we let u = x^3+1, and du = 3x^2 dx, so that x^2dx = (1/3)du, and / / | [x^3+1]^(1/2) x^2 dx = | u^(1/2) (1/3)du / / = (1/3) * (2/3)u^(3/2) + C = (2/9) [x^3+1]^(3/2) + C . Substitution makes the process fairly mechanical so it doesn't require much thought, once you see the appropriate substitution to use, and it also automatically keeps the constants straight. The objective of indefinite integration is to find an antiderivative, and exactly how you do that isn't really important, at least in my opinion. If you can just look at a formula and "see" what the antiderivative is, you have solved the problem. Most of us can't do that, and there are a number of procedures to help, such as the substitution rule, and integration by parts (the integral version of the Product Rule of differentiation). The real theory behind substitution is the Chain Rule, and you can look at the details of substitution as a formal process for helping you see the important parts of the composite functions involved, without worrying about their intrinsic meaning. If you have any questions, please write back and I will try to explain further. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ Date: 12/14/2006 at 13:45:51 From: Reuben Subject: Thank you (Why integration by substitution works) Thank you so much. Your response has made the reasoning behind this SO much clearer to me! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/