Evaluate the following line integral over the curve given by x = t2 , y = t3 , where 1 ≤t ≤ 2.∫ ! !""#𝑑𝑠

To evaluate the line integral $\int \!\! f \, ds$ over the given curve defined by $x = t^2$, $y = t^3$ for $1 \leq t \leq 2$, we need to follow these steps:

1. Break Down the Problem

We need to calculate the line integral of the function $f$ (which is unspecified in the integral prompt and so will be kept general for this solution) over the differential arc length $ds$ along the specified path.

2. Relevant Concepts

The differential arc length $ds$ is calculated using the formula: $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

3. Analysis and Detail

1. Parameterize $dx$ and $dy$:

   • $x = t^2$ implies $dx = 2t \, dt$.
   • $y = t^3$ implies $dy = 3t^2 \, dt$.

2. Calculate $ds$: $\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2$ $\sqrt{\left(2t\right)^2 + \left(3t^2\right)^2} = \sqrt{4t^2 + 9t^4} = t\sqrt{4 + 9t^2}$ Therefore, $ds = t\sqrt{4 + 9t^2} \, dt$

3. Set Up the Integral: The line integral becomes: $\int_{t=1}^{t=2} f(t) \cdot t \sqrt{4 + 9t^2} \, dt$

4. Verify and Summarize

We successfully set up the integral needed to be evaluated. Since $f(t)$ is not defined in the question, the final expression for the line integral is: $\int_{1}^{2} f(t) \cdot t \sqrt{4 + 9t^2} \, dt$

Final Answer

The result of the line integral is: $\int_{1}^{2} f(t) \cdot t \sqrt{4 + 9t^2} \, dt$ where $f(t)$ is the function that needs to be specified for further evaluation.