Respuesta :
Answer:
[tex]\frac{7}{3} + \frac{e^2 - e}{2}[/tex]
Step-by-step explanation:
By definition:
Work done along the path is the line integral along that path denoted as:
Work Done = [tex]\int\limits^C {F} \, dr[/tex]
Note: dr = dx i + dy j
Given that: [tex]F (x,y) = x^2 i + ye^x j[/tex]
F (x, y) dot product with dr = [tex] x^2 dx + ye^x dy[/tex]
Work done = [tex]\int\limits^C {(x^2 dx + ye^x dy)}[/tex] ... Eq 1
Given that C: [tex]y = \sqrt{x-1}[/tex]
[tex]dy = \frac{dx}{2\sqrt{x-1} }[/tex]
Replace the value of y and dy in Eq 1
[tex]Work done = \int\limits^C ({x^2 + \frac{e^x}{2} }) \, dx[/tex]
Limits of x are 1 to 2 respectively
[tex]Work done = \int\limits^2_1 ({x^2 + \frac{e^x}{2} }) \, dx[/tex]
= [tex](\frac{x^3}{3} + \frac{e^x}{2})\limits^2_1[/tex]
Evaluate limits to obtain
Work Done = [tex]\frac{7}{3} + \frac{e^2 - e}{2}[/tex]
