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Show that if $f$ and $g$ are continuous on $[a, b]$ and $[c, d ]$ , respectively, then$$b \int_{a}^{b} \int_{c}^{d}\left[f(x)+g(y) d y d x=(d-c) \int_{a}^{b} f(x) d x\right.$$$$+\int_{a}^{b} \int_{c}^{d} g(y) d y d x=(b-a) \int_{c}^{d} g(y) d y+\int_{c}^{d} \int_{a}^{b} f(x) d x d y$$

$\int_{a}^{b} \int_{c}^{d}[f(x)+g(y)] d y d x=(d-c) \int_{a}^{b} f(x) d x+\int_{a}^{b} \int_{c}^{d} g(y) d y d x=(b-a) \int_{c}^{d} g(y) d y+\int_{c}^{d} \int_{a}^{b} f(x) d x d y$

Calculus 2 / BC

Calculus 3

Chapter 5

Multiple Integration

Section 1

Double Integrals over Rectangular Regions

Integration Techniques

Multiple Integrals

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Okay. This question would like us to derive a formula for this type of double integral if f in gear continuous functions. So we're going to use the linearity of Integral is here to rewrite this as thean a girl from A to B off splitting this up term by term, we get the integral from C to D of f of x d y because our why integral comes first and then are integral of G of way. So we actually don't know anything about this g of why integral of what it could be. But we actually do know how to evaluate this f of X integral because F of X is just a constant in the wide world. So we have actually pull out f of X from the integral because it's just a constant in terms of why. And we know that the integral of one over a line is just the length of the integration window. So we have the integral from a to B of f of x times D minus C plus the integral from cdd of G of why de y and then we're integrating that with respect to X. So now we're going to do the same trick again this time. Distributing are other integral. So now we have pulling our constant out front. Here we get D minus C times the integral from A to B of F of X plus the integral from A to B of the integral from D to see of g of y d Y DX. And I'm Mr DX in this integral here and now we see that we can actually pull out this g of why from our exit a girl. So let's just swap the integration order. So we get D minus C times the integral from a to B of F of X dx and then plus the integral from seated d of the integral from A to B of G of x y or G of y, the x d y and a gun. We could do that because we're just swapping differentials cause we're integrating over a rectangle. We could just use food Beanies theory for that. So now we have d minus C times the integral from a to B of of of x, t x and then well, g of why it's just a constant in this other integral. So we have g of why do y times the integral from A to B dx and again the differentials can also move around. But for completeness, you technically should put the differential in the very end. Sexually will do that. So now you can see by symmetry We're going to get a very similar results as we got our first dinner girl, because the width of the inter girls once again just coming out in front because that's a constant. And now, once I add my differential will have our final answer. And this is what it wanted us to prove. So all we had to do was use linearity.

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