If C is the curve y = x^2/2 + 1 from (0, 1) to (2, 3), then the line integral over C
3x ds =
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To find the line integral of 3x ds over the curve C, you need to follow these steps:
Step 1: Determine the parameterization of the curve C.
The given curve C is defined by the equation y = x^2/2 + 1. To parameterize this curve, we can express x in terms of t, where t is the parameter. Let's assume x = t. Therefore, substituting this into the equation of the curve, we get y = (t^2)/2 + 1.
Step 2: Find the derivative of the parameterization.
To calculate the derivative, we need to differentiate both sides of the equation from Step 1 with respect to t. So, dy/dt = (d/dt)((t^2)/2) + (d/dt)(1), which simplifies to dy/dt = t. Since we assumed x = t, dx/dt = 1.
Step 3: Determine the limits of integration.
The curve C starts at point (0, 1) and goes to point (2, 3). Therefore, we need to find the values of t that correspond to these points. For the starting point, when x = 0, we have t = 0. For the ending point, when x = 2, we have 2 = t. Hence, the limits of integration are t = 0 to t = 2.
Step 4: Calculate the line integral.
The line integral of 3x ds over the curve C can be expressed as ∫(a to b)(3x ds), where a is the lower limit (0) and b is the upper limit (2).
Now, we need to express ds in terms of t. ds represents the length of the curve, which can be calculated using the formula ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt.
Since dx/dt = 1 and dy/dt = t (from Step 2), we have ds = sqrt((1)^2 + (t)^2) dt.
Substituting x = t and ds = sqrt((1)^2 + (t)^2) dt, the line integral becomes ∫(0 to 2)(3t)(sqrt((1)^2 + (t)^2)) dt.
Now, you can solve this integral using the appropriate integration techniques.