# How do you evaluate the line integral, where c is the line segment from (0,8,4) to (6,7,7)?

Let

Hence,

and

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To evaluate the line integral along the line segment from (0,8,4) to (6,7,7), denoted as (C), you first parameterize the line segment using a parameter (t). You can express the parameterization as:

[x(t) = 6t, \quad y(t) = 8 - t, \quad z(t) = 4 + 3t]

Next, you need to find the differential arc length, (ds), along the curve using the formula:

[ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} dt]

Then, you evaluate the line integral of a vector function (F(x, y, z)) along the curve (C) using the formula:

[\int_C F(x, y, z) \cdot ds = \int_a^b F(x(t), y(t), z(t)) \cdot \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} dt]

where (a) and (b) are the initial and final parameter values, respectively.

By substituting the parameterization and (ds) into the line integral formula, you can evaluate the line integral.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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