MIT18_024s11_ChDnotes

MIT18_024s11_ChDnotes - Notes - double integrals. on (Read...

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Notes - on double integrals. (Read 11.1-11.5 of Apostol.) Just as for the case of a single integral, we have the following condition for the existence of a double integral: Theorem 1 (Riemann condition). Suppose f - is defined - on Q = [arb] x [c,d] . Then f - is integrable - Q -- if and only - if given any E > 0, there are step functions s - and t with - s < f < t - on Q, such that - Let A -- be a number. - If these step functions -s - and t satisfy - the further condition that - then A = 11 f. Q The proof is almost identical with the corresponding proof for the single integral. Using this condition, one can readily prove the three basic properties--linearityr additivity, and comparison--for the integral f. We state them as follows: Theorem 2. (a) Suppose f - and g - are integrable - on Q. --- so is cf (x) + dg ; furthermore,
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(b) Let - Q - be subdivided -- into two rectangles Q1 - and Q2. - Then f - is integrable over Q ------ if and only if it integrable over both - Q2; furthermore, (c) - If f g - on Q, and if f - g - are integrable over - then TO prove this theorem, one first verifies these results for step functions (see 11.3), and then uses the Riemann condi- tion to prove them for general integrable functions. The proofs are very similar to those given for the single integral. We give one of the proofs as an illustration. For example, consider the formula lJQ + = JJ, f + I\ Q g, where f g are integrable. We choose step functions sl, s2, tl, t2 such that J s1 C f C tl S2 C g < and such that
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We then find a single partition of Q relative to whi.ch all of sl, s2, tl, t2 are step functions; then sl + s2 and tl + t2 are also step functions relative to this partition. Furthermore, one adds the earlier inequalities to obtain Finally, we compute this computation uses the fact that linearity has already been proved for step functions. ~hus JJQ (f + g) exists. TO calculate this integral, we note that by definition. ' Then here again we use the linearity of the double integral for step functions. It follows from the second half of the Riemann
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conditi.on that (f + g) must equal ;he number Q up to this point, the development of the double integral has been remarkably similar to the development of the single integral. Now things begin to change. We have the following basic questions to answer: (1) Under 11 what conditions does ( f exist? "Q (2) rf f exists, how can one evaluate. it? Q (3) Is there a version of the substitution rule for double integrals? (4) What are the applications of the double integral? We shall deal with questions (l), (Z), and (4) now, postponing question until the next unit. I Let us tackle question first. How can one evaluate the integral if one knows it exists? The answer is that such integrals can almost always be evaluated by repeated one-dimen- sional integration. More precisely, one has the following theorem: Theorem 3 (Fubini theorem). - Let f - be defined - and bounded -- on a rectangle Q = [arb] x [c,d] , - and assume - that f - is integrable - on Q. For each fixed y - in [c,d], assume that the - one-dimensional integral exists. Then the integral A(y) dy exists, , and furthermore, 1
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I proof. We need to show that [ X(y)dy exists and equals the double integral ,b f*
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This note was uploaded on 01/19/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024s11_ChDnotes - Notes - double integrals. on (Read...

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