18.6k views
3 votes
Evaluate the line integral ?c eˣ dx, with the given curve C.

C is the arc of the curve x = y³ from (-1,-1) to (1,1).

User Cedmundo
by
8.1k points

1 Answer

4 votes

The line integral ∫C e^x dx along the curve C, where C is the arc of the curve x = y³ from (-1,-1) to (1,1), is equal to e^(1/4) - e^(-1/4).

The line integral ∫C e^x dx is an expression that represents the cumulative effect of the function e^x along a specified curve C. In this case, the curve C is defined by the arc of the curve x = y³, and it extends from the point (-1,-1) to (1,1).

To evaluate this line integral, we need to parameterize the curve C. Let's express y in terms of a parameter t, such that y = t. Substituting this into the curve equation x = y³ gives x = t³. Now, the limits of integration become from t = -1 to t = 1.

The line integral becomes ∫ from -1 to 1 of e^(t³) * (3t²) dt. To solve this, we recognize that the derivative of t³ is 3t², and the integral becomes ∫ from -1 to 1 of e^(t³) d(t³).

Now, integrating e^(t³) with respect to t³ and evaluating from -1 to 1 gives the result e^(1/4) - e^(-1/4). Therefore, the line integral ∫C e^x dx along the curve C is equal to e^(1/4) - e^(-1/4).

User Takendarkk
by
8.2k points