IntegrationDate: 01/08/98 at 01:30:50 From: B.L.T. Subject: Integration Dr. Math, I was recently doing some Calculus homework and I realized a possible problem in integration. I began to wonder what the integrated area is if the anti-derivative is undefined. I'll show you what I mean: INT(x^-1) from a to b The process I would follow would be to raise the power by 1 and divide by the new power, thus: (x^0)/0 My math knowledge tells me that anything divided by 0 is undefined. Is this true? If so, how does one find the area of the function? Thanks a lot for your time, B.L.T. Date: 01/12/98 at 12:34:42 From: Doctor Joe Subject: Re: Integration Dear B.L.T., I must praise you for your sharp observation regarding the rule that you have told me: Increase power and divide the expresson by the new increased power. But lo and behold! your rule works for only powers of linear factors with the power not equal to -1. Thus, the example that you show precisely falls under this category! Nonetheless, we have a very neat answer. This is classical calculus work as derived by Euler. All derivations are due to Euler and I deserve no glory whatsoever in this work. We first look at the following constant called the natural (Euler) number e (which may just appear somewhere on your Casio Scientific calculators!) e is defined as the limit of the function (1+1/x)^x as x tends to positive infinity. Here, I assume that you have some brief knowledge of limits (at least pictorially you know what it means to be the limit). If you use the calculator, you will discover that e is approximately 2.718. Anyway, we look at the following: d(e^t)/dt = lim (e^(t+h)-e^t)/h h->0 = lim e^t (e^h -1)/h h->0 = e^t lim lim {{(1+1/x)^x}^h-1}/h h->0 x->+inf = e^t lim lim {{(1+1/x)^(xh)}-1}/h x->+inf h->0 = e^t lim lim (1/h)(h + h(xh-1)(1/x)+ ...+ terms in h) x->+inf h->0 = e^t lim lim 1 + (xh-1)/x + terms in h x->+inf h->0 = e^t It follows that d(e^t)/dt = e^t. Now, if x = e^y, it follows from implicit differentiation that 1 = e^y dy/dx. This is the crucial part. Since the exponential function x(y) = e^y is a one-to-one and onto map from [0,inf) to [1,inf), we may define an inverse function denoted by ln such that ln: [1,inf) --> [0,inf) and moreover, ln(e^y) = y and e^(ln x) = x. This ln function is actually none other than the natural log (i.e. log based e). Thus, 1 = e^y dy/dx means dy/dx = e^(-y) = 1/x which further implies that d(ln x)/dx = 1/x. Now, the rest is up to you. -Doctor Joe, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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