Question

Let $b \neq 0$ and for $j=0,1,2,....,n$, let $S_j$ be the area of the region bounded by the y-axis and the curve $xe^{ay}=sin\,by,\frac{j\pi}{b} \leq y \leq \frac{(j+1)\pi}{b}$. 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 1

The sum of the geometric progression is $\frac{\pi(e+1)}{1+\pi^2} \frac{e^{n+1}-1}{e-1}$.

Answer 2

The area $S_j$ of the region bounded by the y-axis ($x=0$) and the curve $xe^{ay}=\sin(by)$ for $\frac{j\pi}{b} \leq y \leq \frac{(j+1)\pi}{b}$ is given by the integral: $S_j = \int_{\frac{j\pi}{b}}^{\frac{(j+1)\pi}{b}} |x| dy$.

From the given curve equation, $x = e^{-ay} \sin(by)$.

Since $e^{-ay}$ is always positive, we need to consider the sign of $\sin(by)$.

For $y \in \left[\frac{j\pi}{b}, \frac{(j+1)\pi}{b}\right]$, the term $by$ lies in the interval $[j\pi, (j+1)\pi]$. In this interval, $\sin(by)$ has a constant sign.

Therefore, $S_j = \left| \int_{\frac{j\pi}{b}}^{\frac{(j+1)\pi}{b}} e^{-ay} \sin(by) dy \right|$.

To evaluate the integral, we use the standard formula $\int e^{Ax} \sin(Bx) dx = \frac{e^{Ax}}{A^2+B^2} (A \sin(Bx) - B \cos(Bx))$.

Here, $A = -a$ and $B = b$.

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

Let $F(y) = \frac{e^{-ay}}{a^2+b^2} (-a \sin(by) - b \cos(by))$.

The definite integral $I_j = \int_{\frac{j\pi}{b}}^{\frac{(j+1)\pi}{b}} e^{-ay} \sin(by) dy = F\left(\frac{(j+1)\pi}{b}\right) - F\left(\frac{j\pi}{b}\right)$.

At the upper limit $y = \frac{(j+1)\pi}{b}$:

$by = (j+1)\pi \implies \sin(by) = 0$ and $\cos(by) = (-1)^{j+1}$.

$F\left(\frac{(j+1)\pi}{b}\right) = \frac{e^{-a\frac{(j+1)\pi}{b}}}{a^2+b^2} (-b (-1)^{j+1}) = \frac{b(-1)^j e^{-a\frac{(j+1)\pi}{b}}}{a^2+b^2}$.

At the lower limit $y = \frac{j\pi}{b}$:

$by = j\pi \implies \sin(by) = 0$ and $\cos(by) = (-1)^j$.

$F\left(\frac{j\pi}{b}\right) = \frac{e^{-a\frac{j\pi}{b}}}{a^2+b^2} (-b (-1)^j)$.

Now, substitute these into $I_j$:

$I_j = \frac{b(-1)^j e^{-a\frac{(j+1)\pi}{b}}}{a^2+b^2} - \left( \frac{-b(-1)^j e^{-a\frac{j\pi}{b}}}{a^2+b^2} \right)$

$I_j = \frac{b(-1)^j}{a^2+b^2} \left( e^{-a\frac{(j+1)\pi}{b}} + e^{-a\frac{j\pi}{b}} \right)$

$I_j = \frac{b(-1)^j}{a^2+b^2} e^{-a\frac{j\pi}{b}} \left( e^{-a\frac{\pi}{b}} + 1 \right)$.

The area $S_j = |I_j|$. Since $b \neq 0$, $a^2+b^2 > 0$, $e^{-a\frac{j\pi}{b}} > 0$, and $e^{-a\frac{\pi}{b}} + 1 > 0$, we have:

$S_j = \frac{|b|}{a^2+b^2} e^{-a\frac{j\pi}{b}} \left( e^{-a\frac{\pi}{b}} + 1 \right)$.

To show that $S_0, S_1, ..., S_n$ are in geometric progression, we find the ratio $\frac{S_{j+1}}{S_j}$:

$\frac{S_{j+1}}{S_j} = \frac{\frac{|b|}{a^2+b^2} e^{-a\frac{(j+1)\pi}{b}} \left( e^{-a\frac{\pi}{b}} + 1 \right)}{\frac{|b|}{a^2+b^2} e^{-a\frac{j\pi}{b}} \left( e^{-a\frac{\pi}{b}} + 1 \right)}$

$\frac{S_{j+1}}{S_j} = \frac{e^{-a\frac{(j+1)\pi}{b}}}{e^{-a\frac{j\pi}{b}}} = e^{-a\frac{\pi}{b}}$.

Since the ratio is constant and independent of $j$, the sequence $S_0, S_1, ..., S_n$ is a geometric progression with common ratio $r = e^{-a\frac{\pi}{b}}$.

Now, we find the sum for $a=-1$ and $b=\pi$.

The common ratio $r = e^{-(-1)\frac{\pi}{\pi}} = e^1 = e$.

The first term $S_0$ (for $j=0$) is:

$S_0 = \frac{|\pi|}{(-1)^2+\pi^2} e^{-(-1)\frac{0\pi}{\pi}} \left( e^{-(-1)\frac{\pi}{\pi}} + 1 \right)$

$S_0 = \frac{\pi}{1+\pi^2} (1) (e+1) = \frac{\pi(e+1)}{1+\pi^2}$.

The sum of a geometric progression with $n+1$ terms (from $j=0$ to $j=n$) is given by the formula:

Sum $= S_0 \frac{r^{n+1}-1}{r-1}$.

Substituting the values of $S_0$ and $r$:

Sum $= \frac{\pi(e+1)}{1+\pi^2} \frac{e^{n+1}-1}{e-1}$.