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Evaluate the given integral.$$\int_{2}^{3} \int_{2 y}^{y^{2}}\left(x^{2} y-y^{2} x-2 x+3 y+1\right) d x d y$$

$$\frac{13897}{168}$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

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for this problem we are asked to evaluate the double integral. Where the into grand is X squared Y Y minus Y squared X minus two. X plus three Y plus one. And we first integrate over X from two Y two Y squared then over why? From 2 to 3? So begin nor to begin, we know that our innermost integral there is going to be an integral over X. So we'll be integrating so we'd have x squared becomes X cubed over three. So we've executed y over three. Then we'd have Y squared X squared over two. Then we have minus two X squared over two or just minus X squared. Then we'd have plus three YX plus X evaluated from two, Y up to y squared. And that is going to be integrated over dy now that or that innermost integral. There can actually be simplified down into the form. See here or rather after evaluating it from endpoint to endpoint in simplifying this can be written as three Y cubed minus uh two Y minus Y squared -5/3. Why? to the power of four -Y to the power of 6/2 Plus Y. to the power of 7/3. And that is being integrated over why? For between two and 3? So integrating Now we would have three Y to the power of 4/4 minus two, Y squared over two. Or just y squared -Y cubed over three minus 5/3 Y to the power of 5/5. So that becomes Y two? Power of 5/3 minus widespread power of 7/7 times one half. So that's minus white power of 7/14. Then we have plus 1/3 times Y to the power of 8/8. So that would become plus Y to the power of 8/8 16 24 Evaluated from 2- three. Well, give us the final result of 13,000, or excuse me, 13,897 13897, Divided by 168.

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