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Evaluate the double integral $\int_{R} \int f(x, y) d A$ over the indicated region $R$, for the given function $f$. $f(x, y)=x^{2} e^{x y}, R$ is the region bounded by $y=0, y=x, x=1$ and $x=2$.

$$\frac{e^{4}-e-3}{2}$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

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Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Evaluate the double integr…

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for this problem we are asked to evaluate the double integral of F. Of X. Y equals X squared to each. The power of xy over the region bounded by Y equals zero. Y equals x X equals one in X equals two. So to begin, we want to plot out our region. We look we can see that our region is the interior the shape here. So we can set this up as a horizontally simple region. So we'll be integrating outside X from 1 to 2 and inside why? From zero up to X. And then we have integral of X squared E R X Y B Y E X. Now integrating first over why is pretty simple affair? Yeah. I'll just note that we would do use substitution U equals E. T. Or a. U equals X times Y. In which case we'll find that the inner integral will become uh X. E to the power of X times Y DX. That's evaluated from 0 to X. So this will then be the integral from 1 to 2 of X. E. To the power of X squared dx. In which case we can now make the U substitution. U equals X squared. So we'd have that D U equals two X dx. So D X equals D. You over to. So this will then become 1/2 times the integral from U. Of one up to you of to of just E. To the power of you do you Which would then be 1/2 times eat the power of you Evaluated from you of one up to you of to Which will then give us 1/2 times now. We plug back in you and then we'd have eat the power of two squared. So that would be e to the power of four minus E. to the power of one. Actually I need to back up here because at this point this actually should have been X times each star of X squared minus one. So we would have to also, so we have um I have here up to this point, the integral from 1 to 2 of exceed the power of X squared. But we would actually have to do integral from 1 to 2 of each power of X squared dx minus Integral from 1 to 2 of XDX. So I'll finish off what I was writing. So we have just 1/2 times each of our four minus E to the power of one. But then integrating uh X we have x squared over two evaluated from 1 to 2. So that would be 4/2, which would be too -1/2. So that would be minus or that would be 3/2. So we would have that, our overall integral would be equal to Uh 1/2. Eat there are 4 -1 to the power of 1 -3

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