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Given the integral $\int_{R} \int d A,$ where $R$ is the region bounded by the $x$ -axis, the line $x=4$ and the line $y=8 x . S$ the region $R$ and determine the value of this integral without doing any integration.

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Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

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Lectures

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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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Hi there. This problem. We were asked to find the area which is above, and our function is X to the fourth minus eight. So above that but beneath the X axis. So it's a draw. Something like this Extra the fourth minus aids just to get a little impression of what we're dealing with. Look, something like this Extra fourth shifting down by eight units. All right, so what we're gonna want is a definite integral as usual. But what we're missing is these two limits of integration. So we need to figure out what they are. In other words, we need to figure out when does extra the fourth minus a equal zero. No solution just would be excellent forthe People's eight. In other words, we want plus or minus the fourth route of eight. We'll just leave it like that for now. Um, next up, we're really ready to go. We can go the definite integral from the negative four through debate all the way until the positive for through debate of our function extra the fourth minus aid the ex to make us the easier. Let's use the fact that we have a symmetrical function with an even function here, so this will be the same instead of taking this whole area. Let's just take the area on the right and then double it. So we'll go on Lee from zero to the four through eight. But then double our final answer. The classic trick that'll make the math little easier for us. Okay. And, uh, fundamental theorem says we can take the anti derivative, which is actually fifth over five minus eight x, and we'll just have to plug in four through debates and then zero. Let's go ahead and do that so that it looks a little nasty here. But we'll take the fourth fruit of eight. All of the fifth. Who are that by five minus eight times the fourth root of AIDS on. We should subtract what we give me it. Plug in zero. But we can see right away by plugging in zero. And for ex both of these terms will become zero. So we can write a minus zero there if we want no need. Really? So I mean, in a sense, this is their final answer. It's a little ugly. Um, will be too times now what? We have here left 2/5 times. It's eight to the 5/4 power minus C. We have eight times eight to the 1/4 so again will have aged to the 5/4 power, right, because it's one plus 1/4 as exponents will give us into the 5/4 so you can actually factor out that eight to the 5/4 and 2/5 minus one, of course, is negative 3/5. So you want to calculate this out on your calculator? You can, but otherwise you can leave it like negative 3/5 times eight to the 5/4 power. And that's your final answer. Um, that's the Integral, by the way, were asked to find, Let's see, since it actually asked us to find the area. The area is self, who probably should take the negative offering areas always considered positive. So if you want, since we know the area was all beneath the X axis, the actual area, we'll just be our number without the minus sign. Okay, now we're done

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