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.

Answer:25/2Step-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] [tex]\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt[/tex] Where P, Q are scalar functions We want to compute [tex]\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy[/tex] 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 [tex]\large C_1,C_2,C_3,C_4[/tex] which represents the sides of the rectangle. [tex]\large C_1[/tex] is the line segment from (0,0) to (5,0) [tex]\large C_2[/tex] is the line segment from (5,0) to (5,1) [tex]\large C_3[/tex] is the line segment from (5,1) to (0,1) [tex]\large C_4[/tex] is the line segment from (0,1) to (0,0) Then [tex]\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}[/tex] 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 [tex]\large C_1[/tex] as r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and r'(t) = (5,0) so [tex]\large \displaystyle\int_{C_1}xydx+x^2dy=0[/tex]  Now the second integral. We parametrize [tex]\large C_2[/tex] as r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and r'(t) = (0,1) so [tex]\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25[/tex] The third integral. We parametrize [tex]\large C_3[/tex] as r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and r'(t) = (-5,0) so [tex]\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2[/tex] The fourth integral. We parametrize [tex]\large C_4[/tex] as r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and r'(t) = (0,-1) so [tex]\large \displaystyle\int_{C_4}xydx+x^2dy=0[/tex] So [tex]\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2[/tex] Now, let us compute the value using Green's theorem. According with this theorem [tex]\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx[/tex] where A is the interior of the rectangle. so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1} We have [tex]\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x[/tex] so [tex]\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2[/tex]