25 - 26 show that the line integral is independent of path and evaluate the integral.\n25. $\\int_{c} 2 x…

25 - 26 show that the line integral is independent of path and evaluate the integral.\n25. $\\int_{c} 2 x e^{-y} d x+(2 y-x^{2} e^{-y}) d y$,\n$c$ is any path from $(1,0)$ to $(2,1)$\n26. $\\int_{c} \\sin y d x+(x \\cos y-\\sin y) d y$,\n$c$ is any path from $(2,0)$ to $(1, \\pi)$

25 - 26 show that the line integral is independent of path and evaluate the integral.\n25. $\\int_{c} 2 x e^{-y} d x+(2 y-x^{2} e^{-y}) d y$,\n$c$ is any path from $(1,0)$ to $(2,1)$\n26. $\\int_{c} \\sin y d x+(x \\cos y-\\sin y) d y$,\n$c$ is any path from $(2,0)$ to $(1, \\pi)$

Answer

Explanation:

Step1: Check independence of path

For a line integral (\int_{C}Pdx + Qdy), if (\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}), the integral is path - independent. For problem 25: (P = 2xe^{-y}), (\frac{\partial P}{\partial y}=- 2xe^{-y}); (Q = 2y-x^{2}e^{-y}), (\frac{\partial Q}{\partial x}=-2xe^{-y}). For problem 26: (P=\sin y), (\frac{\partial P}{\partial y}=\cos y); (Q=x\cos y-\sin y), (\frac{\partial Q}{\partial x}=\cos y).

Step2: Find potential function (f)

Problem 25:

Since (\frac{\partial f}{\partial x}=P = 2xe^{-y}), integrate with respect to (x): (f(x,y)=\int2xe^{-y}dx=x^{2}e^{-y}+g(y)). Differentiate (f) with respect to (y): (\frac{\partial f}{\partial y}=-x^{2}e^{-y}+g^{\prime}(y)). Since (\frac{\partial f}{\partial y}=Q = 2y - x^{2}e^{-y}), then (g^{\prime}(y)=2y), integrate (g(y)=\int2y dy=y^{2}+C). Let (C = 0), so (f(x,y)=x^{2}e^{-y}+y^{2}). Evaluate (f(2,1)-f(1,0)): [ \begin{align*} f(2,1)-f(1,0)&=(2^{2}e^{-1}+1^{2})-(1^{2}e^{0}+0^{2})\ &=\frac{4}{e}+1 - 1\ &=\frac{4}{e} \end{align*} ]

Problem 26:

Since (\frac{\partial f}{\partial x}=P=\sin y), integrate with respect to (x): (f(x,y)=x\sin y+g(y)). Differentiate (f) with respect to (y): (\frac{\partial f}{\partial y}=x\cos y+g^{\prime}(y)). Since (\frac{\partial f}{\partial y}=Q=x\cos y-\sin y), then (g^{\prime}(y)=-\sin y), integrate (g(y)=\cos y + C). Let (C = 0), so (f(x,y)=x\sin y+\cos y). Evaluate (f(1,\pi)-f(2,0)): [ \begin{align*} f(1,\pi)-f(2,0)&=(1\times\sin\pi+\cos\pi)-(2\times\sin0+\cos0)\ &=(0 - 1)-(0 + 1)\ &=-2 \end{align*} ]

Answer:

For problem 25, the value of the line integral is (\frac{4}{e}). For problem 26, the value of the line integral is (-2).