Show that the given line integral is independent of path.Then, evaluate the line integral I by finding a potential function f and applying the Fundamental Theorem of Line Integrals. I=�ç_{(0,0)}^{(1,2)}(x+y)dx+(x-y)dy
If F is a conservative field, then there will be a G such that F = ∇G.
If we let G(x,y) = 1/2 (x^2 + 2xy - y^2) then
∇G = 1/2 (2x dx + 2y dx + 2x dy - 2y dy)
= (x+y)dx + (x-y)dy
So, F is a conservative field and the line integral is path-independent.
So, just evaluate G at the limits of integration
G(1,2) = 1/2 (1+4-4) = 1/2
G(0,0) = 0
So, the integral evaluates to just 1/2
If I recall my vector calculus correctly...
To show that the line integral is independent of the path, we need to use the concept of conservative vector fields. A vector field F is called conservative if it can be expressed as the gradient of a scalar potential function f(x, y). Mathematically, this means that F = ∇f.
If the given line integral of F = (x + y, x - y) is independent of path, it implies that F is a conservative vector field, and we can find a potential function f such that F = ∇f.
To find the potential function f, we integrate each component of F with respect to its corresponding variable:
∫(x + y)dx = ∫x dx + ∫y dx = (1/2)x^2 + xy + C1
∫(x - y)dy = ∫x dy - ∫y dy = xy - (1/2)y^2 + C2
Therefore, the potential function f is given by f(x, y) = (1/2)x^2 + xy + C1, xy - (1/2)y^2 + C2.
Now, we can evaluate the line integral I by applying the Fundamental Theorem of Line Integrals. According to the theorem, if F is a conservative vector field, and f is the potential function, then ∫C F · dr = f(B) - f(A), where C is the curve from point A to point B.
Given that the curve C goes from (0, 0) to (1, 2), we have A = (0, 0) and B = (1, 2). Plugging these values into the potential function f, we get:
f(1, 2) - f(0, 0) = [(1/2)(1)^2 + (1)(2) + C1] - [(1/2)(0)^2 + (0)(0) + C1]
= (1/2) + 2 + C1 - 0 - 0 - C1
= 2.5
Thus, the value of the line integral I is 2.5.