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Question-76826




Question Number 76826 by Master last updated on 30/Dec/19
Answered by john santu last updated on 31/Dec/19
= ∫ _0 ^2  ∫ _0 ^( 1)  ∫ _0 ^( 3)  (2y+z+x) dz dy dx   = ∫ _0 ^2  ∫ _0 ^( 1)  (2yz+(1/2)z^2 +xz)∣_( 0) ^( 3)  dy dx   = ∫ _0 ^( 2)  ∫ _0 ^( 1)  (6y+(9/2)+3x) dy dx  = ∫_0 ^( 2)  (3y^2 +3xy+(9/2)y)∣ _0 ^1  dx  = ∫_0 ^2  (((15)/2)+3x) dx = ((15)/2)x+(3/2)x^2  ∣_( 0) ^( 2)   = 15 + 6 = 21. (done).
$$=\:\int\underset{\mathrm{0}} {\overset{\mathrm{2}} {\:}}\:\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\int\underset{\mathrm{0}} {\overset{\:\mathrm{3}} {\:}}\:\left(\mathrm{2}{y}+{z}+{x}\right)\:{dz}\:{dy}\:{dx}\: \\ $$$$=\:\int\underset{\mathrm{0}} {\overset{\mathrm{2}} {\:}}\:\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\left(\mathrm{2}{yz}+\frac{\mathrm{1}}{\mathrm{2}}{z}^{\mathrm{2}} +{xz}\right)\underset{\:\mathrm{0}} {\overset{\:\mathrm{3}} {\mid}}\:{dy}\:{dx}\: \\ $$$$=\:\int\underset{\mathrm{0}} {\overset{\:\mathrm{2}} {\:}}\:\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\left(\mathrm{6}{y}+\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{3}{x}\right)\:{dy}\:{dx} \\ $$$$=\:\underset{\mathrm{0}} {\overset{\:\mathrm{2}} {\int}}\:\left(\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{xy}+\frac{\mathrm{9}}{\mathrm{2}}{y}\right)\mid\underset{\mathrm{0}} {\overset{\mathrm{1}} {\:}}\:{dx} \\ $$$$=\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\left(\frac{\mathrm{15}}{\mathrm{2}}+\mathrm{3}{x}\right)\:{dx}\:=\:\frac{\mathrm{15}}{\mathrm{2}}{x}+\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \:\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}} {\mid}} \\ $$$$=\:\mathrm{15}\:+\:\mathrm{6}\:=\:\mathrm{21}.\:\left({done}\right).\: \\ $$
Commented by Master last updated on 31/Dec/19
thanks
$$\mathrm{thanks}\: \\ $$

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