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 ratio `(S_(2009))/(S_(2010))` equals :

Options

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

Answer

To solve the problem, we need to find the ratio \( \frac{S_{2009}}{S_{2010}} \) where \( S_j \) is the area bounded by the x-axis and the curve defined by the equation \( y e^x = \sin x \) for \( j\pi \leq x \leq (j+1)\pi \).

### Step-by-Step Solution:

1. **Rearranging the Equation**:
We start with the equation \( y e^x = \sin x \). Rearranging gives us:
\[ y = e^{-x} \sin x \]

2. **Defining 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. **Integration by Parts**:
We will use integration by parts to evaluate the integral \( \int e^{-x} \sin x \, dx \). Let:
\[ u = \sin x \quad \text{and} \quad dv = e^{-x} dx \]
Then,
\[ du = \cos x \, dx \quad \text{and} \quad v = -e^{-x} \]
Applying integration by parts:
\[ \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int -e^{-x} \cos x \, dx \]

4. **Second Integration by Parts**:
Now we need to evaluate \( \int e^{-x} \cos x \, dx \) using integration by parts again:
Let:
\[ u = \cos x \quad \text{and} \quad dv = e^{-x} dx \]
Then,
\[ du = -\sin x \, dx \quad \text{and} \quad v = -e^{-x} \]
Thus,
\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x + \int e^{-x} \sin x \, dx \]

5. **Setting Up the Equation**:
Let \( I = \int e^{-x} \sin x \, dx \). We have:
\[ I = -e^{-x} \sin x + e^{-x} \cos x - I \]
Rearranging gives:
\[ 2I = -e^{-x} (\sin x + \cos x) \]
Therefore,
\[ I = -\frac{1}{2} e^{-x} (\sin x + \cos x) \]

6. **Evaluating the Integral**:
Now we evaluate \( S_j \):
\[ S_j = \left[-\frac{1}{2} e^{-x} (\sin x + \cos x)\right]_{j\pi}^{(j+1)\pi} \]

7. **Calculating \( S_{2009} \) and \( S_{2010} \)**:
For \( S_{2009} \):
\[ S_{2009} = -\frac{1}{2} \left( e^{-(2009+1)\pi} (\sin((2009+1)\pi) + \cos((2009+1)\pi)) - e^{-2009\pi} (\sin(2009\pi) + \cos(2009\pi)) \right) \]
Simplifying using \( \sin(n\pi) = 0 \) and \( \cos(n\pi) = (-1)^n \):
\[ S_{2009} = -\frac{1}{2} \left( e^{-2010\pi}(-1) - e^{-2009\pi}(1) \right) \]
\[ S_{2009} = \frac{1}{2} (e^{-2010\pi} + e^{-2009\pi}) \]

For \( S_{2010} \):
\[ S_{2010} = -\frac{1}{2} \left( e^{-(2010+1)\pi} (\sin((2010+1)\pi) + \cos((2010+1)\pi)) - e^{-2010\pi} (\sin(2010\pi) + \cos(2010\pi)) \right) \]
\[ S_{2010} = \frac{1}{2} (e^{-2011\pi} + e^{-2010\pi}) \]

8. **Finding the Ratio**:
Now we find the ratio:
\[ \frac{S_{2009}}{S_{2010}} = \frac{\frac{1}{2} (e^{-2010\pi} + e^{-2009\pi})}{\frac{1}{2} (e^{-2011\pi} + e^{-2010\pi})} \]
Simplifying this gives:
\[ \frac{S_{2009}}{S_{2010}} = \frac{e^{-2010\pi} + e^{-2009\pi}}{e^{-2011\pi} + e^{-2010\pi}} = \frac{e^{\pi} + 1}{1 + e^{\pi}} = e^{\pi} \]

### Final Answer:
\[ \frac{S_{2009}}{S_{2010}} = e^{\pi} \]