1. a) equation of the line :
y - intercept = -7
so, it will pass through point (0, -7)
and if we plug the value of x as 5, we get
so, it will pass through point (5, -5) too
now, just plot the points (0 , -7) and (5 , -5) and join them.
2. b) equation of line is :
here, y - intercept = 5
so the line passes through point (0 , 5)
now, Plugging the value of x = 1 we get :
so, the given line passes through point (1 , 2)
plotting the points, we can get our required line.
Answer:
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