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Q: Calculus: Improper Integrals ( Answered ,   2 Comments )
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 Subject: Calculus: Improper Integrals Category: Science > Math Asked by: megahog-ga List Price: \$2.50 Posted: 06 Jul 2005 13:12 PDT Expires: 05 Aug 2005 13:12 PDT Question ID: 540640
 `Is 1/(the cubed root of "x") convergent or divergent on the interval of -1 to 8?` Clarification of Question by megahog-ga on 06 Jul 2005 14:33 PDT `- This is an INTEGRAL!` Request for Question Clarification by mathtalk-ga on 06 Jul 2005 19:25 PDT ```Hi, megahog-ga: I understand that you are asking about a particular improper integral, ie. whether it is convergent or divergent. This terminology suggests you are studying either the Riemann integral or the Riemann-Stieltjes integral. Although either of these would give the same Answer regarding the improper integral as a sum of limits of proper integrals in this case, it would be helpful to know more about the context of your Question. In fact there are many definitions of integration, and they differ on some "badly behaved" functions, so knowing which type of definition you are working will clarify things. regards, mathtalk-ga``` Clarification of Question by megahog-ga on 07 Jul 2005 00:29 PDT ```Riemann integral - I believe. Basically, this integral was given in a test I took recently. The course os Calculus II - so it couldnt be that advanced. I basically just want to know if the "area" of 1/the cubed root of x converges to a specific value or if it diverges to an infinite value. The notation of the integral involves the elongated "s" if that helps. The limits are from -1 to 8.```
 Subject: Re: Calculus: Improper Integrals Answered By: mathtalk-ga on 07 Jul 2005 14:43 PDT Rated:
 ```Hi, megahog-ga: The definition and theory of Riemann integrals is commonly presented first for the case of a continuous function integrated over a finite (bounded) interval. An improper integral in this context is then some sort of case where either the integrand (function) is not continuous over the interval of integration, the interval is not bounded, or both. In your problem the interval of integration [-1,8] is certainly bounded, so the only issue that arises is the one point at with the function to be integrated is discontinous, i.e. 1/x^(1/3) is discontinuous at x = 0. For simplicity we will write the function to be integrated as: f(x) = x^(-1/3) since one over the cube root of x is also expressed as the minus one-third power of x. The circumflex ^ is often used in math "texting" to mean a superscript, esp. as in this case an exponent. Many examples treat a discontinuity occuring at an endpoint of the interval of integration, but here the discontinuity lies strictly inside that interval. The approach to be taken as far as finding out whether the improper integral converges or diverges is to break the interval of integration into two parts, so that the discontinuity lies at the upper limit of [-1,0] and at the lower limit of [0,8]. If both "parts" of the integral converge, then we say the original improper integral also converges. Otherwise we conclude that it diverges: INTEGRAL f(x) dx OVER [-1,8] = INTEGRAL f(x) dx OVER [-1,0] + INTEGRAL f(x) dx OVER [0,8] Then we use a standard approach to each piece, taking a limit of integrals whose upper (resp. lower) limits of integration approach 0, thus reducing the question of the convergence of the improper integral to a limit of "proper" integrals: INTEGRAL f(x) dx OVER [-1,0] = LIMIT ( INTEGRAL f(x) dx OVER [-1,u] ) u -> 0- (i.e. the limit as u tends to zero from below) INTEGRAL f(x) dx OVER [0,8] = LIMIT ( INTEGRAL f(x) dx OVER [v,8] ) v -> 0+ (i.e. the limit as v tends to zero from above) To evaluate the integrals inside the LIMIT's, we can use the fact that f(x) is continuous on intervals [-1,u] and [v,8] and has an "anti-derivative" there: f(x) = dF(x)/dx where F(x) = (3/2) * x^(2/3) (Check my differentiation of F(x) by using the Power Rule for derivatives. One can obtain other anti-derivatives of f(x) by adding any constant to F(x).) The Fundamental Theorem of Calculus then gives us the definite integrals of f(x) in terms of this anti-derivative: INTEGRAL f(x) dx OVER [-1,u] = F(u)-F(-1) = (3/2)*(u^(2/3)) - (3/2)*((-1)^(2/3)) = (3/2)*(u^(2/3)) - (3/2)*1 INTEGRAL f(x) dx OVER [v,8] = F(8)-F(v) = (3/2)*(8^(2/3)) - (3/2)*(v^(2/3)) = (3/2)*4 - (3/2)*(v^(2/3)) Due to the positive exponent 2/3rds in these expressions, the limits as u,v tend to zero both exist, and both are obtained by "substituting" 0 for u and v since the "2/3rds power" function is continuous there: LIMIT ( INTEGRAL f(x) dx OVER [-1,u] ) = 0 - (3/2) = -3/2 u -> 0- LIMIT ( INTEGRAL f(x) dx OVER [v,8] ) = (3/2)*4 - 0 = 6 v -> 0+ That is, both "pieces" of the original improper integral converge, and so it converges as well. The value of the original improper integral is furthermore the sum of these two values above, namely -3/2 + 6 = 9/2. regards, mathtalk-ga```
 megahog-ga rated this answer: `Very pleased with your elaborated response - very thorough and assuring.`

 ```Did you try pluggin the equation into your calculator and calculating the integral that way? I did it and after a minute or so, I obtained 4.499997522.... Calc2 is certainly the most difficult of all the calc courses.```
 ```Hi, 80d-ga: Your numerical approximation shows good agreement with the exact value 4.5. regards, mathtalk-ga```