101. Find the area bounded by the curve
x = cos t, y = e
t
, 0 ≤ t ≤
π
2
and the lines y = 1 and
x = 0

To find the area bounded by the given parametric curve \(x = \cos t, y = e^t\) for \(0 \leq t \leq \frac{\pi}{2}\), and the lines \(y = 1\) and \(x = 0\), we need to set up an appropriate integral. We will follow these steps:

1. **Describe the region**: The region is bounded by the curve \(x = \cos t, y = e^t\), the vertical line \(x = 0\), and the horizontal line \(y = 1\).

2. **Find the parametric expressions for the area**: The area under a parametric curve \(x = f(t), y = g(t)\) from \(t = a\) to \(t = b\) is given by:
   \[
   A = \int_{a}^{b} g(t) \frac{dx}{dt} \, dt
   \]
   Here, \(g(t) = e^t\) and \(f(t) = \cos t\).

3. **Calculate \( \frac{dx}{dt} \)**:
   \[
   \frac{dx}{dt} = -\sin t
   \]

4. **Set up the integral**:
   The area is given by:
   \[
   A = \int_{0}^{\frac{\pi}{2}} e^t (-\sin t) \, dt
   \]
   Simplifying, this becomes:
   \[
   A = -\int_{0}^{\frac{\pi}{2}} e^t \sin t \, dt
   \]

5. **Evaluate the integral**:
   We need to solve \( \int_{0}^{\frac{\pi}{2}} e^t \sin t \, dt \). We can use integration by parts, where we set \(u = \sin t\) and \(dv = e^t dt\), leading to \(du = \cos t dt\) and \(v = e^t\).

   Applying integration by parts:
   \[
   \int e^t \sin t \, dt = e^t \sin t - \int e^t \cos t \, dt
   \]

   Repeating integration by parts for \( \int e^t \cos t \, dt \), set \(u = \cos t\) and \(dv = e^t dt\), leading to \(du = -\sin t dt\) and \(v = e^t\):
   \[
   \int e^t \cos t \, dt = e^t \cos t - \int e^t (-\sin t) \, dt
   \]
   This results in:
   \[
   \int e^t \sin t \, dt = e^t \sin t - (e^t \cos t - \int e^t \sin t \, dt)
   \]
   Combining terms gives:
   \[
   2 \int e^t \sin t \, dt = e^t (\sin t - \cos t)
   \]
   Therefore,
   \[
   \int e^t \sin t \, dt = \frac{e^t (\sin t - \cos t)}{2}
   \]

6. **Evaluate at the bounds**:
   \[
   A = -\left[\frac{e^t (\sin t - \cos t)}{2}\right]_{0}^{\frac{\pi}{2}}
   \]
   Calculating this, we get:
   \[
   A = -\frac{1}{2} \left[e^{\frac{\pi}{2}} \left(\sin \frac{\pi}{2} - \cos \frac{\pi}{2}\right) - e^0 \left(\sin 0 - \cos 0\right)\right]
   \]
   \[
   = -\frac{1}{2} \left[e^{\frac{\pi}{2}} (1 - 0) - 1 (0 - 1)\right]
   \]
   \[
   = -\frac{1}{2} \left(e^{\frac{\pi}{2}} + 1\right)
   \]
   Thus, the area is:
   \[
   A = \frac{1 + e^{\frac{\pi}{2}}}{2}
   \]

The area bounded by the curve and the given lines is \(\frac{1 + e^{\frac{\pi}{2}}}{2}\).

Would you like more details or have any questions?

Here are five related questions:
1. How would you find the area for different bounds of \(t\)?
2. What if the parametric curve was given as a different function of \(t\)?
3. How does integration by parts help in solving parametric integrals?
4. What are other methods to find the area under a parametric curve?
5. How do you handle cases where the curve intersects the boundary lines?

**Tip**: For parametric curves, always ensure to correctly set up the integral based on the derivatives of \(x\) and \(y\).

Find the area bounded by the curve x = cos t, y = e^t, 0 ≤ t ≤ π/2 and the lines y = 1 and x = 0.

Learn how to calculate the area bounded by the parametric curve x = cos t, y = e^t between 0 ≤ t ≤ π/2 and the lines y = 1 and x = 0. This solution uses integration by parts and explores parametric curves, ideal for advanced high school or early college students.

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