A curve is represented parametrically by the equations `x=e^(1)cost andy=e^(1) sin t,` where t is a parameter. Then, If `F(t)=int(x+y)dt,` then the value of `F((pi)/(2))-F(0)` is

Options

A. 1
B. -1
C. `e^(pi//2)`
D. 0

Answer

To solve the problem, we will follow these steps:

1. **Identify the Parametric Equations**: We have the parametric equations given as:
\[ x = e^t \cos t \]
\[ y = e^t \sin t \]

2. **Define the Function \( F(t) \)**: The function \( F(t) \) is defined as:
\[ F(t) = \int (x + y) \, dt \]
Substituting the expressions for \( x \) and \( y \):
\[ F(t) = \int (e^t \cos t + e^t \sin t) \, dt \]

3. **Combine the Terms**: We can factor out \( e^t \):
\[ F(t) = \int e^t (\cos t + \sin t) \, dt \]

4. **Integration by Parts or Recognizing the Form**: We can use the formula for integrating \( e^x (f(x) + f'(x)) \):
\[ \int e^t (\cos t + \sin t) \, dt = e^t (\sin t - \cos t) + C \]
Therefore, we have:
\[ F(t) = e^t (\sin t - \cos t) + C \]

5. **Evaluate \( F\left(\frac{\pi}{2}\right) \) and \( F(0) \)**:
- For \( F\left(\frac{\pi}{2}\right) \):
\[ F\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \left(\sin\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right)\right) + C = e^{\frac{\pi}{2}} (1 - 0) + C = e^{\frac{\pi}{2}} + C \]
- For \( F(0) \):
\[ F(0) = e^0 (\sin(0) - \cos(0)) + C = 1(0 - 1) + C = -1 + C \]

6. **Calculate \( F\left(\frac{\pi}{2}\right) - F(0) \)**:
\[ F\left(\frac{\pi}{2}\right) - F(0) = \left(e^{\frac{\pi}{2}} + C\right) - \left(-1 + C\right) \]
The \( C \) terms cancel out:
\[ = e^{\frac{\pi}{2}} + 1 \]

Thus, the final answer is:
\[ F\left(\frac{\pi}{2}\right) - F(0) = e^{\frac{\pi}{2}} + 1 \]