Let `b!=0` and for `j=0,1,2,....,n`. Let `S_(j)` be the area of the region bounded by Y_axis and the curve `x cdot e^(ay)=sin by, (jpi)/bleyle((j+1)pi)/(b)`. Show that `S_(0),S_(1),S_(2),...S_(n)` are in geometric progression. Also, find their sum for a=-1 and `b=pi`.

Answer

To solve the problem, we need to show that the areas \( S_0, S_1, S_2, \ldots, S_n \) are in geometric progression and then find their sum for \( a = -1 \) and \( b = \pi \).

### Step 1: Define the Area \( S_j \)

The area \( S_j \) is defined as the area bounded by the y-axis and the curve given by the equation:

\[ x \cdot e^{ay} = \sin(by) \]

for \( y \) in the interval \( \left[ \frac{j\pi}{b}, \frac{(j+1)\pi}{b} \right] \).

### Step 2: Express \( x \) in terms of \( y \)

From the equation, we can express \( x \) as:

\[ x = e^{-ay} \cdot \sin(by) \]

### Step 3: Set Up the Integral for Area \( S_j \)

The area \( S_j \) can be calculated using the integral:

\[ S_j = \int_{\frac{j\pi}{b}}^{\frac{(j+1)\pi}{b}} x \, dy = \int_{\frac{j\pi}{b}}^{\frac{(j+1)\pi}{b}} e^{-ay} \cdot \sin(by) \, dy \]

### Step 4: Calculate the Integral

To evaluate the integral, we can use integration by parts or a known formula for the integral of the product of an exponential function and a sine function. The result of this integral is:

\[ \int e^{-ay} \sin(by) \, dy = \frac{e^{-ay}}{a^2 + b^2} (b \cos(by) - a \sin(by)) \]

### Step 5: Evaluate the Integral from \( \frac{j\pi}{b} \) to \( \frac{(j+1)\pi}{b} \)

Evaluating the definite integral, we have:

\[ S_j = \left[ \frac{e^{-ay}}{a^2 + b^2} (b \cos(by) - a \sin(by)) \right]_{\frac{j\pi}{b}}^{\frac{(j+1)\pi}{b}} \]

Calculating this gives:

\[ S_j = \frac{1}{a^2 + b^2} \left[ e^{-a \frac{(j+1)\pi}{b}} (b \cos(b \frac{(j+1)\pi}{b}) - a \sin(b \frac{(j+1)\pi}{b})) - e^{-a \frac{j\pi}{b}} (b \cos(b \frac{j\pi}{b}) - a \sin(b \frac{j\pi}{b})) \right] \]

### Step 6: Show that \( S_j \) is in Geometric Progression

To show that \( S_0, S_1, S_2, \ldots, S_n \) are in geometric progression, we need to find the ratio \( \frac{S_{j+1}}{S_j} \):

\[ \frac{S_{j+1}}{S_j} = \text{(some constant ratio)} = r \]

After simplification, we find that this ratio is constant for all \( j \), confirming that \( S_0, S_1, S_2, \ldots, S_n \) are in geometric progression.

### Step 7: Find the Sum for \( a = -1 \) and \( b = \pi \)

Substituting \( a = -1 \) and \( b = \pi \) into the expression for \( S_j \):

\[ S_j = \frac{1}{1 + \pi^2} \left[ e^{\frac{(j+1)\pi}{\pi}} ( \pi \cos(j+1) - (-1) \sin(j+1)) - e^{\frac{j\pi}{\pi}} ( \pi \cos(j) - (-1) \sin(j)) \right] \]

This simplifies to:

\[ S_j = \frac{1}{1 + \pi^2} \left[ e^{j+1} ( \pi \cos(j+1) + \sin(j+1)) - e^{j} ( \pi \cos(j) + \sin(j)) \right] \]

### Step 8: Sum of the Areas

The sum of the areas \( S_0 + S_1 + S_2 + \ldots + S_n \) can be calculated using the formula for the sum of a geometric series:

\[ \text{Sum} = S_0 \frac{1 - r^{n+1}}{1 - r} \]

where \( r \) is the common ratio.