Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2) (a) directly (b) using Green's Theorem

Answer:a) [tex] \frac{2}{3} [/tex]b) [tex] \frac{2}{3} [/tex]Step-by-step explanation:a) The first part requires that we use line integral to evaluate directly.The line integral is[tex] \int_C xydx + {x}^{2} {y}^{3} dy[/tex]where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2) The boundary of integration is shown in the attachment.Our first line integral is [tex]L_1 = \int_ {(0,0)}^{(1,0)} xydx + {x}^{2} {y}^{3} dy[/tex]The equation of this line is y=0, x varies from 0 to 1.When we substitute y=0 every becomes zero.[tex] \therefore \: L_1 =0[/tex]Our second line integral is [tex]L_2 = \int_ {(1,0)}^{(1,2)} xydx + {x}^{2} {y}^{3} dy[/tex]The equation of this line is:[tex]x = 0 \implies \: dx = 0[/tex]y varies from 1 to 2.We substitute the boundary and the values to get:[tex]L_2 = \int_ {1}^{2}1 \cdot y(0) + {1}^{2} \cdot \: {y}^{3} dy[/tex][tex]L_2 = \int_ {1}^2 {y}^{3} dy = \frac{8}{3} [/tex]The 3rd line integral is:[tex]L_3 = \int_ {(1,2)}^{(0,0)} xydx + {x}^{2} {y}^{3} dy[/tex]The equation of this line is [tex]y = 2x \implies \: dy = 2dx[/tex]x varies from 0 to 1.We substitute to get:[tex]L_3 = \int_ {1}^{0} x \cdot \: 2xdx + {x}^{2} {(2x)}^{3}(2 dx)[/tex][tex]L_3 = \int_ {1}^{0} 8 {x}^{5} + 2 {x}^{2} dx = - 2[/tex]The value of the line integral is [tex]L = L_1 + L_2 + L_3[/tex][tex]L = 0 + \frac{8}{3} + - 2 = \frac{2}{3} [/tex]b) The second part requires the use of Green's Theorem to evaluate:[tex] \int_C xydx + {x}^{2} {y}^{3} dy[/tex]Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.This is given by:[tex] \int_C \: Pdx +Q \: dy = \int \int_ R \: Q_y - P_x \: dA[/tex][tex]\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int \int_ R \: 3 {x}^{2} {y}^{2} - y \: dA[/tex]We choose our region of integration parallel to the y-axis.[tex]\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \int_ 0^{2x} \: 3 {x}^{2} {y}^{2} - y \: dydx[/tex][tex]\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: {x}^{2} {y}^{3} - \frac{1}{2} {y}^{2} |_ 0^{2x} dx[/tex][tex]\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: 8{x}^{5} - 2 {x}^{2} dx = \frac{2}{3} [/tex]