SATHEE: Area Question 35

Area Question 35

Question 35

Let $b \neq 0$ and for $j = 0, 1, 2 \dots, n$ . If $S_{j}$ is the area of the region bounded by the $Y$ -axis and the curve $x e^{a y} = \sin b y, \frac{j π}{b} \leq y \leq \frac{(j + 1) π}{(b)}$ . Then, show that

$S_{0}, S_{1}, S_{2}, \dots, S_{n}$ are in geometric progression. Also, find their sum for $a = - 1$ and $b = π$ .

$(2001, 5$ M)

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Solution:

Given, $x = (\sin b y) e^{- a y}$

Now,

$- 1 \leq \sin$ by $\leq 1$

$\Rightarrow - e^{- a y} \leq e^{- a y} \sin b y \leq e^{- a y}$

$\Rightarrow - e^{- a y} \leq x \leq e^{- a y}$

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, $S_{j} = \int_{j π / b}^{(j + 1) π / b} \sin b y \cdot e^{- a y} d y$

$Missing or unrecognized delimiter for \left$

For $a = - 1$ and $b = π$

$\begin{aligned} S_{j} & = \frac{π \cdot e^{\frac{1}{π} \cdot π_{j}}}{(1 + π^{2)}} e^{\frac{1}{π} \cdot π} + 1 = \frac{π \cdot e^{j}}{(1 + π^{2)}} (1 + e) \Rightarrow \sum_{j = 0}^{n} S_{j} & = \frac{π \cdot (1 + e)}{(1 + π)^{2}} \sum_{j = 0}^{n} e^{j} = \frac{π (1 + e)}{(1 + π^{2})} (e^{0} + e^{1} + \dots + e^{n}) & = \frac{π (1 + e)}{(1 + π^{2})} \cdot \frac{(e^{n + 1} - 1)}{e - 1} \end{aligned}$

Area Question 35

Question 35

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Area Question 35

Question 35

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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