Area Question 35
Question 35
- Let $b \neq 0$ and for $j=0,1,2 \ldots, n$. If $S_{j}$ is the area of the region bounded by the $Y$-axis and the curve $x e^{a y}=\sin b y, \frac{j \pi}{b} \leq y \leq \frac{(j+1) \pi}{(b)}$. Then, show that
$S_{0}, S_{1}, S_{2}, \ldots, S_{n}$ are in geometric progression. Also, find their sum for $a=-1$ and $b=\pi$.
$(2001,5$ M)
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Solution:
- Given, $x=(\sin b y) e^{-a y}$
Now,
$-1 \leq \sin$ by $\leq 1$
$\Rightarrow \quad-e^{-a y} \leq e^{-a y} \sin b y \leq e^{-a y}$
$\Rightarrow \quad-e^{-a y} \leq x \leq e^{-a y}$
In this case, if we take $a$ and $b$ positive, the values $-e^{-a y}$ and $e^{-a y}$ become left bond and right bond of the curve and due to oscillating nature of sin $b y$, it will oscillate between $x=e^{-a y}$ and $x=-e^{-a y}$
Now, $\quad S_{j}=\int_{j \pi / b}^{(j+1) \pi / b} \sin b y \cdot e^{-a y} d y$
$$ \begin{aligned} & \text { since, } I=\int \sin b y \cdot e^{-a y} d y \ & I=\frac{-e^{-a y}}{a^{2}+b^{2}} \quad(a \sin b y+b \cos b y) \ & \therefore \quad S_{j}=\frac{-1}{a^{2}+b^{2}} e^{\frac{-a(j+1) \pi}{b}} \ & {a \sin (j+1) \pi+b \cos (j+1) \pi} \ & -e^{\frac{-a j \pi}{b}}(a \sin j \pi+b \cos j \pi) \ & S_{j}=\mid-\frac{1}{a^{2}+b^{2}}\left[e^{-\frac{a}{b}(j+1) \pi}\left{0+b(-1)^{j+1}\right}\right. \ & \left.-e^{-a j \pi / b}\left{0+b(-1)^{j}\right}\right] \ & =\frac{b(-1)^{j} e^{-\frac{a}{b} j \pi}}{a^{2}+b^{2}} e^{-\frac{a}{b} \pi}+1 \ & {\left[\because(-1)^{j+2}=(-1)^{2}(-1)^{j}=(-1)^{j}\right]} \ & =\frac{b e^{-\frac{a}{b} j \pi}}{a^{2}+b^{2}} e^{-\frac{a}{b} \pi}+1 \ & b e^{-\frac{a}{b} j \pi} e^{\frac{-a \pi}{b}}+1 \ & \therefore \quad \frac{S_{j}}{S_{j-1}}=\frac{a^{2}+b^{2}}{b e^{-\frac{a}{b}(j-1) \pi} e^{\frac{-a \pi}{b}}+1}=\frac{e^{-\frac{a}{b} j \pi}}{e^{-\frac{a}{b}(j-1) \pi}} \ & a^{2}+b^{2} \ & =e^{-\frac{a}{b} \pi}=\text { constant } \end{aligned} $$
For $a=-1$ and $b=\pi$
$$ \begin{aligned} S_{j} & =\frac{\pi \cdot e^{\frac{1}{\pi} \cdot \pi_{j}}}{\left(1+\pi^{2)}\right.} e^{\frac{1}{\pi} \cdot \pi}+1=\frac{\pi \cdot e^{j}}{\left(1+\pi^{2)}\right.}(1+e) \ \Rightarrow \quad \sum_{j=0}^{n} S_{j} & =\frac{\pi \cdot(1+e)}{(1+\pi)^{2}} \sum_{j=0}^{n} e^{j}=\frac{\pi(1+e)}{\left(1+\pi^{2}\right)}\left(e^{0}+e^{1}+\ldots+e^{n}\right) \ & =\frac{\pi(1+e)}{\left(1+\pi^{2}\right)} \cdot \frac{\left(e^{n+1}-1\right)}{e-1} \end{aligned} $$
