show that the line integral is independent of path. ∫_c 2xe^(-y)dx + (2y - x^2e^(-y))dy, c is any path from…

show that the line integral is independent of path. ∫_c 2xe^(-y)dx + (2y - x^2e^(-y))dy, c is any path from (1, 0) to (5, 1) the functions 2xe^(-y) and 2y - x^2e^(-y) have continuous first - order derivatives on ℝ^2 and ∂/∂y(2xe^(-y)) = = ∂/∂x(2y - x^2e^(-y)), so f(x, y) = ( )i + (2y - x^2e^(-y))j is a conservative vector field by the theorem given below, hence the line integral is independent of path. theorem: let f = pi + qj be a vector field on an open simply - connected region d. suppose that p and q have continuous first - order partial derivatives and ∂p/∂y = ∂q/∂x throughout d. then f is conservative. evaluate the integral.

show that the line integral is independent of path. ∫_c 2xe^(-y)dx + (2y - x^2e^(-y))dy, c is any path from (1, 0) to (5, 1) the functions 2xe^(-y) and 2y - x^2e^(-y) have continuous first - order derivatives on ℝ^2 and ∂/∂y(2xe^(-y)) = = ∂/∂x(2y - x^2e^(-y)), so f(x, y) = ( )i + (2y - x^2e^(-y))j is a conservative vector field by the theorem given below, hence the line integral is independent of path. theorem: let f = pi + qj be a vector field on an open simply - connected region d. suppose that p and q have continuous first - order partial derivatives and ∂p/∂y = ∂q/∂x throughout d. then f is conservative. evaluate the integral.

Answer

Explanation:

Step1: Calcular $\frac{\partial P}{\partial y}$

Dado $P = 2xe^{-y}$, entonces $\frac{\partial P}{\partial y}=2x(-e^{-y})=- 2xe^{-y}$

Step2: Calcular $\frac{\partial Q}{\partial x}$

Dado $Q = 2y - x^{2}e^{-y}$, entonces $\frac{\partial Q}{\partial x}=-2xe^{-y}$

Step3: Encontrar la función potencial $f(x,y)$

Integrar $P = 2xe^{-y}$ con respecto a $x$: $f(x,y)=\int 2xe^{-y}dx=x^{2}e^{-y}+g(y)$. Derivar $f(x,y)$ con respecto a $y$: $\frac{\partial f}{\partial y}=-x^{2}e^{-y}+g'(y)$. Como $\frac{\partial f}{\partial y}=Q = 2y - x^{2}e^{-y}$, entonces $g'(y)=2y$. Integrando $g'(y)$ con respecto a $y$ obtenemos $g(y)=y^{2}+C$. Tomamos $C = 0$, así $f(x,y)=x^{2}e^{-y}+y^{2}$

Step4: Evaluar el integral de línea

El integral de línea $\int_{C}2xe^{-y}dx+(2y - x^{2}e^{-y})dy=f(5,1)-f(1,0)$ $f(5,1)=5^{2}e^{-1}+1^{2}=\frac{25}{e}+1$ $f(1,0)=1^{2}e^{0}+0^{2}=1$ Entonces $\int_{C}2xe^{-y}dx+(2y - x^{2}e^{-y})dy=\frac{25}{e}+1 - 1=\frac{25}{e}$

Answer:

$\frac{25}{e}$