For `j =0,1,2…n ` let `S _(j)` be the area of region bounded by the x-axis and the curve `ye ^(x)=sin x ` for `j pi le x le (j +1) pi`
The value of `sum _(j =0)^(oo) S_(j)` equals to :

Options

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

Answer

To solve the problem, we need to find the value of the sum \( \sum_{j=0}^{\infty} S_j \), where \( S_j \) is the area of the region bounded by the x-axis and the curve \( y e^x = \sin x \) for \( j\pi \leq x \leq (j+1)\pi \).

### Step-by-Step Solution:

1. **Rewrite the equation**:
The given equation can be rewritten as:
\[ y = e^{-x} \sin x \]

2. **Determine the area \( S_j \)**:
The area \( S_j \) can be expressed as:
\[ S_j = \int_{j\pi}^{(j+1)\pi} e^{-x} \sin x \, dx \]

3. **Use integration by parts**:
To solve the integral \( \int e^{-x} \sin x \, dx \), we will use integration by parts. Let:
- \( u = \sin x \) and \( dv = e^{-x} dx \)
- Then, \( du = \cos x \, dx \) and \( v = -e^{-x} \)

Applying integration by parts:
\[ \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int -e^{-x} \cos x \, dx \]
This leads to:
\[ \int e^{-x} \sin x \, dx = -e^{-x} \sin x + \int e^{-x} \cos x \, dx \]

4. **Integrate \( \int e^{-x} \cos x \, dx \)**:
Again, we apply integration by parts:
- Let \( u = \cos x \) and \( dv = e^{-x} dx \)
- Then, \( du = -\sin x \, dx \) and \( v = -e^{-x} \)

This gives:
\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x + \int e^{-x} \sin x \, dx \]

5. **Set up the equation**:
Let \( I = \int e^{-x} \sin x \, dx \). Then we have:
\[ I = -e^{-x} \sin x - e^{-x} \cos x + I \]
Rearranging gives:
\[ 2I = -e^{-x} (\sin x + \cos x) \]
Thus:
\[ I = -\frac{1}{2} e^{-x} (\sin x + \cos x) \]

6. **Evaluate the definite integral**:
Now we can evaluate \( S_j \):
\[ S_j = \int_{j\pi}^{(j+1)\pi} e^{-x} \sin x \, dx = \left[-\frac{1}{2} e^{-x} (\sin x + \cos x)\right]_{j\pi}^{(j+1)\pi} \]

7. **Substituting the limits**:
Evaluating at the limits:
\[ S_j = -\frac{1}{2} \left( e^{-(j+1)\pi} (\sin((j+1)\pi) + \cos((j+1)\pi)) - e^{-j\pi} (\sin(j\pi) + \cos(j\pi)) \right) \]
Since \( \sin(n\pi) = 0 \) and \( \cos(n\pi) = (-1)^n \):
\[ S_j = -\frac{1}{2} \left( e^{-(j+1)\pi} (0 - (-1)^{j+1}) - e^{-j\pi} (0 + (-1)^j) \right) \]
Simplifying gives:
\[ S_j = \frac{1}{2} \left( e^{-j\pi} - e^{-(j+1)\pi} \right) \]

8. **Sum the series**:
Now we sum \( S_j \):
\[ \sum_{j=0}^{\infty} S_j = \sum_{j=0}^{\infty} \frac{1}{2} \left( e^{-j\pi} - e^{-(j+1)\pi} \right) \]
This is a telescoping series, which simplifies to:
\[ \sum_{j=0}^{\infty} S_j = \frac{1}{2} \left( 1 \right) = \frac{1}{2} \]

9. **Final result**:
The final value of the sum \( \sum_{j=0}^{\infty} S_j \) is:
\[ \sum_{j=0}^{\infty} S_j = \frac{1}{2} \cdot \frac{1}{1 - e^{-\pi}} = \frac{1 + e^{-\pi}}{2(1 - e^{-\pi})} \]