Question

[CALCULUS III - DOUBLE INTEGRALS] Evaluate \[\int_{0}^{1}\int_{y-1}^{0}e^{x+y}dxdy\].

Answer 1

\[\int_{y-1}^{0}e^{x+y}dx=e^{x+y}|_{y-1}^{0}=e^{(0)+y}-(e^{(y-1)+y})\]\[=e^y-e^{2y-1}\]

Answer 2

\[=e^y(1-e^{y-1})\]

Answer 3

\[\int_{0}^{1}e^y-e^{2y-1}dy\]

Answer 4

Currently I am approaching it by solving the first integral, solving the second integral, then evaluating the addition.

But I feel like there is some property I am missing that could allow me to avoid all of that, notice how the limits change.

Answer 5

I did a wolfram computation and have verified that they do cancel out, the first integral is c and the second integral is -c, so you obtain

c + -c = 0

Now rather than evaluate each integral, what property or observation am I missing that will allow me to realize the symmetry that allows them to cancel out

Answer 6

reference: http://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx

Answer 7

properties of double integrals: http://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx

Answer 8

Anybody capable and willing to help provide analysis as to how these integrals evaluated and added together equate to zero?

Answer 9

limits flipped gives\[e^{2y-1}-e^y\]

Answer 10

You can factor out the negative to obtain the same integral \[e^y-e^{2y-1}\]

Answer 11

So you are taking \[\int_{0}^{1}u\ du+(-\int_{0}^{1}u\ du)\]\[\int_{0}^{1}u\ du-\int_{0}^{1}u\ du=0\]

Thanks all