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evulate the following double integral [[ 3x-y dy dx where [ is integral sign please help tommowro is exam
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\[\iint(3x-y)dydx.\]First, integrate with respect to y, holding x constant:\[\int(3xy-\frac{1}{2}y^2+c_1)dx.\]Then integrate with respect to x:\[\frac{3}{2}x^2y-\frac{1}{2}xy^2+c_1x+c_2.\]
you say holding y as constant then y u increase y in first integrating 1/3 y2
hmm thats good thanku...
do you treat y as a constant on the the second integration?
Yes because we're integrating with respect to x.
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Also, altafhpatel, with indefinite integrals it's easy to forget including the constants of integration; you can see above that they're fairly important!
Thank you ma'am.
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