Improper integrals These are a special kind of limit. An improper integral is one where either the interval of integration is infinite, or else it includes a singularity of the function being integrated. Examples of the first kind are: 1 1 x2 x e dx and dx 0 - Examples of the second kind: 2 1 3 1 0 x dx and tan x dx 4 The second of these is subtle because the singularity of tan x occurs in the interior of the interval of integration rather than at one of the endpoints. Same method No matter which kind of improper integral (or combination of improper integrals) we are faced with, the method of dealing with them is the same: b e dx means the same thing as lim e dx x x b 0 0 Calculate the limit! What is the value of this limit (and hence, of the improper integral e x dx )? A. 0 0 B. 1 C. e b D. e E. Another improper integral 1 Recall that dx arctan x C. 1 x 2 1 What is the value of dx ? - 1 x 2 A. 0 D. B. 4 E. C. 2 The integral you just worked represents all of the area between the x-axis and the graph of 1 2 1 x Area between the x- axis and the graph The other type... For improper integrals of the other type, we make the same kind of limit definition: 1 1 1 1 0 x dx is defined to be lim a 0 a x dx. Another example: What is the value of this limit, in 1 1 other words, what is 0 x dx ? A. 0 B. 1 C. 2 D. 2 E. A divergent improper integral It is possible that the limit used to define the improper integral fails to exist or is infinite. In this case, the improper integral is said to diverge . If the limit does exist, then the improper integral converges. For example: 1 1 x a 0 0 dx lim ln 1 ln a so this improper integral diverges. One for you: 2 3 Does tan( x)dx converge or diverge? 4 A. Converge B. Diverge Sometimes it is possible... to show that an improper integral converges without actually evaluating it: 1 1 Since 4 4 for all x 0, we have that x x7 x b b 1 1 1 x 4 x 7 dx x 4 dx 3 3b3 for all b 1. 1 1 1 So the limit of the first integral must be finite as b goes to infinity, because it increases as b does but is bounded above (by 1/3). A puzzling example... Consider the surface obtained by rotating the graph of y = 1/x for x > 1 around the x-axis: Let’s calculate the volume contained inside the surface: V dx b 1 b cubic units. 1 2 x lim 1 1 What about the surface area? This is equal to... 2 1 SA 2 f ( x) 1 f ' ( x) dx 2 1 4 dx 1 1 x x This last integral is difficult (impossible) to evaluate directly, but it is easy to see that its integrand is bigger than that of the divergent integral 2 1 x dx Therefore it, too is divergent, so the surface has infinite surface area. This surface is sometimes called "Gabriel's horn" -- it is a surface that can be "filled with water" but not "painted".
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