Evaluate the line integral by the two following methods. xy dx + x2 dy c is counterclockwise around the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) (a) directly -25 incorrect: your answer is incorrect. (b) using green's theorem 50 incorrect: your answer is incorrect.

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

Where P, Q are scalar functions

We want to compute

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths which represents the sides of the rectangle.

is the line segment from (0,0) to (5,0)

is the line segment from (5,0) to (5,1)

is the line segment from (5,1) to (0,1)

is the line segment from (0,1) to (0,0)

Then

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

 Now the second integral. We parametrize as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

The third integral. We parametrize as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

The fourth integral. We parametrize as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

So

Now, let us compute the value using Green's theorem.

According with this theorem

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

so

450 is the final answer! : )

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