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Question Which property of double integrals should be applied as a logical first step to evaluate , (ze+y) dA over the region {(z, y)|0 <1, 0g(.v) for (z,v) in R then UR (z,v)dA > [[nR 9(z,v}dA. O (z.v) is integrable over the rectangular region Rand m (z.v)< M then mx A(R)< ffo f(z.vidA < Mx A(R).

Question Which property of double integrals should be applied as a logical first step to evaluate , (ze+y) dA over the region {(z, y)|0 <1, 0g(.v) for (z,v) in R then UR (z,v)dA > [[nR 9(z,v}dA. O (z.v) is integrable over the rectangular region Rand m (z.v)< M then mx A(R)< ffo f(z.vidA < Mx A(R).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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If f(z, v) and g(z, 9) are integrable over the rectangular region R, then the sum f(z, 9) +g(z,y) is integrable and DR F(z,9) + g(z,9) dA = R (z,v)dA + ORg(z, v)dA. %3D If f(z,y) is integrable over the rectangular region Rand e is a constant, then ef(z,v) is integrable and OR f(z, v)dA cDR f(z,v)dA. Oif f(z,y) is integrable over the rectangular region Rand S and Tare subregions of Rsuch that R SUTand SnT=0 except an overlap on the boundaries, then fR f(z,y)dA - [, f(z,v)dA + [, f(z,v)dA. O if f(z,y) and g(z, y) are integrable over the rectangular region Rand f(z,y)>9(z,y) for (z, y) in R then OR (z,v)dA > R 9(z, v)dA. If f(z,y) is integrable over the rectangular region Rand m< f(z.v) <M, then mx A(R)< fR {z,v)dA <M x A(R). Assume /(z,y) is integrable over the rectangular region R in the case where f(z,y) can be factored as a product of a function g(z) ofzoniy and a function A(y) of yonly, then over the region R-(7,v)a <z <b, e<y<d}, the double integral can be written as , fr.u)dA ()dz) (h(y}dy}. MORE INSTDIICTION SURMIT
Transcribed Image Text:If f(z, v) and g(z, 9) are integrable over the rectangular region R, then the sum f(z, 9) +g(z,y) is integrable and DR F(z,9) + g(z,9) dA = R (z,v)dA + ORg(z, v)dA. %3D If f(z,y) is integrable over the rectangular region Rand e is a constant, then ef(z,v) is integrable and OR f(z, v)dA cDR f(z,v)dA. Oif f(z,y) is integrable over the rectangular region Rand S and Tare subregions of Rsuch that R SUTand SnT=0 except an overlap on the boundaries, then fR f(z,y)dA - [, f(z,v)dA + [, f(z,v)dA. O if f(z,y) and g(z, y) are integrable over the rectangular region Rand f(z,y)>9(z,y) for (z, y) in R then OR (z,v)dA > R 9(z, v)dA. If f(z,y) is integrable over the rectangular region Rand m< f(z.v) <M, then mx A(R)< fR {z,v)dA <M x A(R). Assume /(z,y) is integrable over the rectangular region R in the case where f(z,y) can be factored as a product of a function g(z) ofzoniy and a function A(y) of yonly, then over the region R-(7,v)a <z <b, e<y<d}, the double integral can be written as , fr.u)dA ()dz) (h(y}dy}. MORE INSTDIICTION SURMIT
Question Which property of double integrais should be applied as a logical first step to evaluate , (ze+y) dA over the region (z, v)|0 <S1, 0<y< 2)? R= Select the correct answer below Oif f(z,v) and g(z, 9) are integrable over the rectangular region R, then the sum f(z, y) + g(z, y) is integrable and DR/(z.v) + g(z, )dA - [R f(z,v)dA + fn9(z,v)dA If f(x,v) is integrable over the rectangular region Rand e is a constant, then ef(z,v) is integrable and If f(z,y) is integrable over the rectangular region Rand S and Tare subregions of R such that R SUT and SnT o except an overlap on the boundaries then , /(r,vlLA ff, f(z.v)dA , f(z.v)4A /G.v) and g(, v) are integrabe over the rectangular reg on Rand /(2.v)> g(.0) for (z.y) in R then If f(z,u) is integrable over the rectangular reglon Rand mf(z,v)< M then m x A(R) < fle f(z.v)dA Mx A(R).
Transcribed Image Text:Question Which property of double integrais should be applied as a logical first step to evaluate , (ze+y) dA over the region (z, v)|0 <S1, 0<y< 2)? R= Select the correct answer below Oif f(z,v) and g(z, 9) are integrable over the rectangular region R, then the sum f(z, y) + g(z, y) is integrable and DR/(z.v) + g(z, )dA - [R f(z,v)dA + fn9(z,v)dA If f(x,v) is integrable over the rectangular region Rand e is a constant, then ef(z,v) is integrable and If f(z,y) is integrable over the rectangular region Rand S and Tare subregions of R such that R SUT and SnT o except an overlap on the boundaries then , /(r,vlLA ff, f(z.v)dA , f(z.v)4A /G.v) and g(, v) are integrabe over the rectangular reg on Rand /(2.v)> g(.0) for (z.y) in R then If f(z,u) is integrable over the rectangular reglon Rand mf(z,v)< M then m x A(R) < fle f(z.v)dA Mx A(R).
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