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Let's do another problem.

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Very similar to the last one, but with a subtle difference.

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And that subtle difference will make a big difference.

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Let's say we take the line integral over some curve c--

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I'll define the curve in a second-- of x squared plus y

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squared dx plus 2xy dy-- and this might look very familiar.

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This was very similar to what we saw last time, except last

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time we had a closed line integral.

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This is not a closed line integral.

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And our curve, c, the parameterization is x is

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equal to cosine of t, y is equal to sine of t.

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So far-- it looks like sit.

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Let me write sine of t-- so far, it looks very similar to

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the closed line integral example we did in the last

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video, but instead of t going from 0 to 2 pi, we're going

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to have t go from 0 to pi.

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t is greater than or equal to 0, is less than or equal to pi.

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So now we're essentially, our path-- if I were to draw it on

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the x-y plane-- so that is my y-axis, that is my x-axis.

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So now our path isn't all the way around the unit circle.

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Our path-- our curve c now-- just starts at t is equal to 0.

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You can imagine t is almost the angle.

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t is equal to 0, and we're going to go all the way to pi.

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So that's what our path is right now, in this example.

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So it's not a curved path.

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It's not a closed path.

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So we can't just show that f, in this example-- and we're

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going to re-look at what f looks like-- hey, if that's a

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conservative vector field, if it's a closed loop that equals

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0, this isn't a closed loop.

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So we can't apply that.

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But let's see if we can apply some of our other tools.

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So like we saw in the last video, this might look a

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little bit foreign to you.

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But if you say that f is equal to that times i, plus that

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times j, then it might look a little bit more familiar.

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If we say that f of xy-- the vector field f is equal to x

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squared plus y squared times i plus 2xy times j and dr-- I

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don't even have to look at this right now. dr, you can always

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write it as dx times i plus dy times j.

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You'll immediately see, if you take the dot product of these 2

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things, if you take f dot dr-- they're both vector valued,

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vector valued differential, vector valued field, or vector

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valued function-- if you take f dot dr, you'll get

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this right here.

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You'll get what we have inside of the interval.

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You'll get that right there, right?

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That times that-- you take the product of the i terms--

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that times that is equal to that, and add it to the

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product of the j terms.

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2xy times dy.

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Write like that.

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So our integral, we can rewrite it as this.

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Along this curve of f dot dr, where this is our f.

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Now, we still might want to ask ourselves, is this

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a conservative field?

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Or does it have a potential?

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Is f equal to the gradient of some function, capital F?

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I guess I could write the gradient like that, because

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it creates a vector.

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This is a vector, too.

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Is this true?

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And we saw in the last video, it is.

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I'll redo it a little bit fast this time.

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Because if this is true, we can't say this is a closed

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loop and say, oh, it's just going to be equal to 0.

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But if this is true, then we know that this-- that the

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integral is path independent.

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And we'll know that this is going to be equal to capital F,

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if we say that t is going from-- well, in this case t is

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going from 0 to pi-- we could say that this is going to be

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equal to capital F of pi minus capital F of 0.

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Or if we want to write it in terms of x and y-- because f is

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going to be a function of x and y-- we could write-- and this

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right here, these are t's.

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We could also write that this is equal to f of x of pi, y of

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pi, minus f of x of 0, y of 0.

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That's what I mean when I say f of pi.

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If we were to write f purely as a function of t.

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But we know that this capital F is going to be a function.

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It's a scalar function defined on xy.

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So we could say f of x of pi, y of pi.

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These are the t's now.

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These are all equivalent things.

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So if it is path dependent, we can find our f.

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We can just evaluate this thing by just taking our

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f, evaluating it at these two points.

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At this point, and at that point right there.

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Because it would be path independent.

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If this is a conservative, if this has a potential function,

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if this is the gradient of another scalar field, then this

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is a conservative vector field, and its line integral

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is path independent.

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It's only dependent on that point and that point.

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So let's see if we can find our f.

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So I'm going to do exactly what we did in the last video.

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If you watch that last video, it might be a

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little bit monotonous.

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But I'll do it a little bit faster here.

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So we know that the partial of f with respect to x is going to

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have to be equal to this right here.

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So that's x squared plus y squared.

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Which tells us, if we take the antiderivative, with respect to

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x, then f of xy is going to have to be equal to x to the

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third over 3 plus xy squared-- right? y squared is just a

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constant in terms of x-- plus f of y.

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There might be some function of y that, when you take

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the partial with respect to x, it just disappears.

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And then we know that the partial of f with respect to

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y has got to be equal to that thing or that thing.

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We're saying that this is the gradient of f.

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So this has to be the partial with respect to y.

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2xy.

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And you might want to watch the other video.

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I go through this just a little bit slower in that one.

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So the antiderivative of this with respect to y-- so we get f

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of xy-- would be equal to xy squared plus some

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function of x.

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Now we did this in the last video.

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These 2 things have to be the same thing, in order for the

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gradient of capital F to be lowercase f.

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And we have xy squared, xy squared.

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We have a function of x, we have a function purely of x.

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And then we don't have a function purely of y here,

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so this thing right here must be 0.

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So we've solved.

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Our capital F of xy must be equal to x to the 3

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over 3 plus xy squared.

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So we know that lowercase f is definitely conservative.

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It is path independent.

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It has its potential.

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It is the gradient of this thing right here.

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And so to solve our integral-- this was a 0-- to solve our

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integral, we just have to figure out x of pi, y

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of pi, x of 0, y of 0.

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Evaluate the bullet points, and then subtract the 2.

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So let's do that.

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So x was cosine of t, y is sine of t.

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Let me rewrite it down here.

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So x is equal to cosine of t.

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y is equal to sine of t.

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So x of 0 is equal to cosine of 0, which is equal to 1.

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x of pi is equal to cosine of pi, which is equal to minus 1.

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y of 0 is sine of 0, which is 0.

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y of pi, which is equal to sine of pi, which is equal to 0.

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So f of x of pi, y of pi-- this is the same thing,

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so let me rewrite this.

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Our integral is simplified to-- our integral along that path of

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f dot dr-- is going to be equal to capital F of x of pi.

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x of pi is minus 1.

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y of pi is equal to 0.

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Minus capital F of x of 0 is 1, comma y of 0 is 0.

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And so what is this equal to?

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Just remember, this right here is the same thing

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is that right there.

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That is x of pi.

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That is y of pi.

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That term right there.

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You can imagine this whole f of minus 1, 0-- that's the same

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thing as f of pi, if you think in terms of just t.

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That could be a little confusing, so I want

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to make that clear.

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So this is just straightforward to evaluate.

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What is f of minus 1, 0?

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x is minus 1. y is 0.

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So it's going to be minus 1 to the third power-- right,

170
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that's our x-- over 3.

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So it's minus 1/3.

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It's going to be minus 1/3 plus minus 1 times 0 squared.

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So that's just going to be a 0.

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In both cases, the y is 0.

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So this term is going to disappear.

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So we can ignore that.

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And then we have minus f of 1, comma 0.

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We put a 1 here.

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1 to the third over 3.

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That is 1/3 plus 1 times 0 squared.

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That's just 0.

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So this is going to be equal to minus 1/3.

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Minus 1/3 is equal to minus 2/3.

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And we're done.

185
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And once again, because this is a conservative vector field,

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and it's path independent, we really didn't have to mess with

187
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the cosine of t's and sines of t's when we actually took

188
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our antiderivative.

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We just have to find the potential function and evaluate

190
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it at the 2 end points to get the answer of our integral, of

191
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our line integral, minus 2/3.

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