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Evaluate the triple itnegral $\int_{0}^{1} \int_{x}^{2 x} \int_{0}^{2 x+3 y}(x+2 y-3 z) d z d y d x$.

$$-19 / 2$$

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 integral from 0 to 3 of the integral. From 0 to 2 of the integral from 0 to 1 of X plus Y plus Z said Dx DZ Dy. So we begin by evaluating our inner integral which we can see we are integrating along X. So we take Z and Y and hold them constant and evaluate or integrate along X. So we'd have X squared over two plus xy Plus zero X evaluated from 0 to 1. Dessert. Dy So this will then become the integral from 0 to 3 of the integral from 0 to 2 of now we plug in X equals one. So this will become 1/2 plus why? Plus 0 -0 -0 0. So we then have our next step would be to integrate along Z ed. So we have the integral from 0 to 3 of Z over to plus y zed plus Z squared over two Evaluated from 0 to 2 dy. Which will then give us well let's see here's it over to will become 2/2. So it becomes uh just one. We have one. Then we have plus two, It would be plus two. Y. Then we'd have plus two squared over two. So plus 4/2 or just plus two. Dy So this will become the integral from 0 to 3 of three plus two. Y dy. Which is then going to be equal to three Y plus two. Y squared over two or just +33. Y plus Y squared evaluated from 0 to 3. So this will become then three times three or nine plus three squared. So we have nine plus nine, giving us 18 as our final result.

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