Double Integral with Negative Exponent

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The discussion revolves around evaluating the double integral ∫4to5 ∫1to2 (1x + y)−2 dy dx. Participants express confusion regarding the negative exponent in the integrand. One suggested approach is to treat the negative two as an exponent, leading to the substitution u = x + y, with du = dy, to simplify the integration process. Another participant notes that an antiderivative of -2 with respect to y is -2y, indicating a potential method for integration. The conversation emphasizes the importance of correctly interpreting the negative exponent for successful integration.
Loppyfoot
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Homework Statement



∫4to5 ∫1to2 (1x + y)−2 dy dx





The Attempt at a Solution



I am confused about what to do with this negative 2.

Any ideas?
 

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Loppyfoot said:

Homework Statement



∫4to5 ∫1to2 (1x + y)−2 dy dx





The Attempt at a Solution



I am confused about what to do with this negative 2.

Any ideas?

An antiderivative of -2 with respect to y is -2y. Or if that -2 is supposed to be an exponent, use the X2 button: (x + y)-2. If that is what you mean then let

u = x + y, du = dy

to proceed.
 

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