Let $b \neq 0$ and for $j=0, 1, 2, \dots, n$, let $S_{j}$ be the area of the region bounded by the $y$-axis and the curve $x e^{a y}=\sin b y, \frac{j \pi}{b} \leq y \leq \frac{(j+1) \pi}{b}$. Show that $S_{0}, S_{1}, S_{2}, \dots, S_{n}$ are in geometric progression. Also, find their sum for $a=-1$ and $b=\pi$.

### Analyzing the Setup The curve is defined by the equation $x = e^{-ay} \sin(by)$. This function represents an oscillating wave modulated by an exponential envelope. To find the area of the regions $S_j$ bounded by this curve and the $y$-axis, we integrate the absolute value of $x$ with respect to $y$. We define the $j$-th region as: $$S_j = \int_{\frac{j\pi}{b}}^{\frac{(j+1)\pi}{b}} |e^{-ay} \sin(by)| \, dy$$ ### The Transformation To simplify the integral, we perform the substitution $t = by$, which implies $dy = \frac{dt}{b}$. The limits of integration transform from $y \in [\frac{j\pi}{b}, \frac{(j+1)\pi}{b}]$ to $t \in [j\pi, (j+1)\pi]$. The integral becomes: $$S_j = \frac{1}{b} \int_{j\pi}^{(j+1)\pi} e^{-at/b} |\sin t| \, dt$$ To relate $S_j$ to the first region $S_0$, we substitute $t = j\pi + u$, where $du = dt$. The limits for $u$ become $0$ to $\pi$. ### The Geometric Progression Using the property $|\sin(j\pi + u)| = |\sin u| = \sin u$ for $u \in [0, \pi]$, we rewrite the integral as: $$S_j = \frac{1}{b} \int_{0}^{\pi} e^{-a(j\pi + u)/b} \sin u \, du$$ Factoring out the constant term $e^{-aj\pi/b}$, we obtain: $$S_j = e^{-aj\pi/b} \left( \frac{1}{b} \int_{0}^{\pi} e^{-au/b} \sin u \, du \right)$$ The expression in the parentheses is exactly $S_0$. Thus, we have a geometric progression $S_j = S_0 \cdot r^j$, where the common ratio is $r = e^{-a\pi/b}$. ### Final Calculation Given $a = -1$ and $b = \pi$, the common ratio simplifies to: $$r = e^{-(-1)\pi/\pi} = e^1 = e$$ We evaluate $S_0$ using the standard integral formula $\int e^{Au} \sin(Bu) \, du = \frac{e^{Au}}{A^2+B^2}(A \sin Bu - B \cos Bu)$: $$S_0 = \frac{1}{\pi} \int_{0}^{\pi} e^{u/\pi} \sin u \, du = \frac{\pi(e+1)}{1+\pi^2}$$ The sum of the first $n+1$ regions (from $j=0$ to $n$) is given by the geometric series sum formula $S = S_0 \frac{r^{n+1}-1}{r-1}$. Substituting our values, we arrive at the final result: **$$S = \frac{\pi(1+e)(e^{n+1}-1)}{(1+\pi^2)(e-1)}$$**

JEE Advanced 2001

LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: Let $b \neq = 0$ and for $j = 0, 1, 2, \dots, n$ , let $S_{j}$ be the area of the region bounded by the $y$ -axis and the curve $x e^{a y} = sin b y, \frac{j π}{b} \leq y \leq \frac{( j + 1 ) π}{b}$ . Show that $S_{0}, S_{1}, S_{2}, \dots, S_{n}$ are in geometric progression. Also, find their sum for $a = - 1$ and $b = π$ .

Q: Wait, why do we use the absolute value $|x|$? Isn't the area just the integral of the function?

Great question! Remember, the integral $\int f(x) dx$ gives the *signed* area. If the curve dips below the axis, the integral becomes negative. But 'area' in geometry is a physical quantity—it must be positive. Since our curve $x = e^{-ay} \sin(by)$ oscillates, it crosses the $y$-axis. To get the total area, we must integrate the magnitude $|x|$.

Q: Why does the number of terms in the sum become $n+1$?

This is a classic trap! We are summing $S_j$ for $j = 0, 1, 2, \dots, n$. If you count the integers from $0$ to $n$, you have $n$ numbers plus the zero itself, giving you $n+1$ terms. Always double-check your indices—it's the difference between a correct answer and a silly mistake.

Progressive Hints

Hint 1

Visualize the curve

x = e^{- a y} sin (b y)

as an oscillating wave trapped within an exponential envelope. The key to this problem isn't brute-force integration, but recognizing that the area

S_{j}

is defined by the periodicity of the sine function. Ask yourself: how does the area of the

j

-th region relate to the area of the very first region,

S_{0}

Hint 2

You need to perform a change of variables. Let

t = b y

. This transforms the integral into a form where the limits are multiples of

π

. Then, use the substitution

t = j π + u

to shift the limits of every integral back to the interval

[0, π]

. This will reveal the 'DNA' of the first region inside every subsequent region.

Hint 3

Start by setting up the integral for

S_{j}

S_{j} = \int_{\frac{j π}{b}}^{\frac{( j + 1 ) π}{b}} ∣ e^{- a y} sin (b y) ∣ d y

. After your substitutions, you should arrive at

S_{j} = \frac{1}{b} \int_{0}^{π} e^{- a (j π + u) / b} sin (u) d u

. Factor out the terms independent of

u

to isolate

S_{0}

Visualized Solution

Visualizing the Regions $S_{j}$

Given curve: $x e^{a y} = sin b y \Rightarrow x = e^{- a y} sin b y$
The regions $S_{j}$ are bounded by the $y$ -axis ( $x = 0$ ) and the curve.
The interval for $S_{j}$ is $y \in [\frac{j π}{b}, \frac{( j + 1 ) π}{b}]$ .

Setting up the Integral for $S_{j}$

Area is the integral of $∣ x ∣$ with respect to $y$ .
$S_{j} = \int_{\frac{j π}{b}}^{\frac{( j + 1 ) π}{b}} ∣ e^{- a y} sin b y ∣ d y$
Since $e^{- a y} > 0$ for all real $y$ , we can simplify the absolute value.
$S_{j} = \int_{\frac{j π}{b}}^{\frac{( j + 1 ) π}{b}} e^{- a y} ∣ sin b y ∣ d y$

Substitution $t = b y$

Let's simplify the argument of the sine function.
Substitute $b y = t \Rightarrow d y = \frac{d t}{b}$ .
When $y = \frac{j π}{b}$ , $t = j π$ .
When $y = \frac{( j + 1 ) π}{b}$ , $t = (j + 1) π$ .
$S_{j} = \frac{1}{b} \int_{j π}^{(j + 1) π} e^{- a t / b} ∣ sin t ∣ d t$

Shifting the Limits to $0$ to $π$

To relate $S_{j}$ to $S_{0}$ , we shift the limits to start from $0$ .
Let $t = j π + u \Rightarrow d t = d u$ .
When $t = j π$ , $u = 0$ . When $t = (j + 1) π$ , $u = π$ .
$S_{j} = \frac{1}{b} \int_{0}^{π} e^{- a (j π + u) / b} ∣ sin (j π + u) ∣ d u$

Simplifying the Integrand

Analyze the sine term: $∣ sin (j π + u) ∣ = ∣ (- 1)^{j} sin u ∣ = ∣ sin u ∣$ .
Since $u \in [0, π]$ , $sin u \geq 0$ , so $∣ sin u ∣ = sin u$ .
Expand the exponential: $e^{- a (j π + u) / b} = e^{- aj π / b} \cdot e^{- a u / b}$ .
$S_{j} = \frac{1}{b} e^{- aj π / b} \int_{0}^{π} e^{- a u / b} sin u d u$

Proving the Geometric Progression

Notice the integral part: $\frac{1}{b} \int_{0}^{π} e^{- a u / b} sin u d u$ is exactly $S_{0}$ .
Therefore, $S_{j} = (e^{- aπ / b})^{j} S_{0}$ .
This shows that $S_{0}, S_{1}, S_{2}, \dots$ form a Geometric Progression.
The common ratio is $r = e^{- aπ / b}$ .

Substituting $a = - 1$ and $b = π$

The problem asks for the sum when $a = - 1$ and $b = π$ .
Let's find the specific common ratio $r$ .
$r = e^{- (- 1) π / π} = e^{1} = e$ .
The first term becomes $S_{0} = \frac{1}{π} \int_{0}^{π} e^{u / π} sin u d u$ .

Integration Formula for $S_{0}$

We need to evaluate $S_{0} = \frac{1}{π} \int_{0}^{π} e^{u / π} sin u d u$ .
Use the standard integration formula:
$\int e^{A x} sin B x d x = \frac{e ^{A x}}{A ^{2} + B ^{2}} (A sin B x - B cos B x)$
Comparing with our integral, we have $A = \frac{1}{π}$ and $B = 1$ .

Evaluating $S_{0}$

Apply the formula: $\int_{0}^{π} e^{u / π} sin u d u = [\frac{e ^{u / π}}{( 1/ π ) ^{2} + 1} (\frac{1}{π} sin u - cos u)]_{0}^{π}$
Upper limit ( $π$ ): $\frac{e ^{1}}{1/ π ^{2} + 1} (0 - (- 1)) = \frac{π ^{2} e}{1 + π ^{2}}$
Lower limit ( $0$ ): $\frac{e ^{0}}{1/ π ^{2} + 1} (0 - 1) = \frac{- π ^{2}}{1 + π ^{2}}$
Integral value = $\frac{π ^{2} ( e + 1 )}{1 + π ^{2}}$
Finally, $S_{0} = \frac{1}{π} \cdot \frac{π ^{2} ( e + 1 )}{1 + π ^{2}} = \frac{π ( e + 1 )}{1 + π ^{2}}$

Sum of the Geometric Progression

We need the sum $S = S_{0} + S_{1} + S_{2} + \dots + S_{n}$ .
This is a G.P. with $n + 1$ terms.
The sum formula is $S = S_{0} \frac{r ^{n + 1} - 1}{r - 1}$ .
Substitute $S_{0} = \frac{π ( e + 1 )}{1 + π ^{2}}$ and $r = e$ .

Final Result

Sum $S = \frac{π ( e + 1 )}{1 + π ^{2}} \cdot \frac{e ^{n + 1} - 1}{e - 1}$
Rearranging the terms for the final elegant form:
$S = \frac{π ( 1 + e ) ( e ^{n + 1} - 1 )}{( 1 + π ^{2} ) ( e - 1 )}$
This matches the required answer perfectly.

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

The curve is defined by the equation

x = e^{- a y} sin (b y)

. This function represents an oscillating wave modulated by an exponential envelope.

To find the area of the regions

S_{j}

bounded by this curve and the

y

-axis, we integrate the absolute value of

x

with respect to

y

. We define the

j

-th region as:

S_{j} = \int_{\frac{j π}{b}}^{\frac{( j + 1 ) π}{b}} ∣ e^{- a y} sin (b y) ∣ d y

The Transformation

To simplify the integral, we perform the substitution

t = b y

, which implies

d y = \frac{d t}{b}

. The limits of integration transform from

y \in [\frac{j π}{b}, \frac{( j + 1 ) π}{b}]

t \in [j π, (j + 1) π]

The integral becomes:

S_{j} = \frac{1}{b} \int_{j π}^{(j + 1) π} e^{- a t / b} ∣ sin t ∣ d t

To relate

S_{j}

to the first region

S_{0}

, we substitute

t = j π + u

, where

d u = d t

. The limits for

u

become

0

π

The Geometric Progression

Using the property

∣ sin (j π + u) ∣ = ∣ sin u ∣ = sin u

for

u \in [0, π]

, we rewrite the integral as:

S_{j} = \frac{1}{b} \int_{0}^{π} e^{- a (j π + u) / b} sin u d u

Factoring out the constant term

e^{- aj π / b}

, we obtain:

S_{j} = e^{- aj π / b} (\frac{1}{b} \int_{0}^{π} e^{- a u / b} sin u d u)

The expression in the parentheses is exactly

S_{0}

. Thus, we have a geometric progression

S_{j} = S_{0} \cdot r^{j}

, where the common ratio is

r = e^{- aπ / b}

Final Calculation

Given

a = - 1

and

b = π

, the common ratio simplifies to:

r = e^{- (- 1) π / π} = e^{1} = e

We evaluate

S_{0}

using the standard integral formula

\int e^{A u} sin (B u) d u = \frac{e ^{A u}}{A ^{2} + B ^{2}} (A sin B u - B cos B u)

S_{0} = \frac{1}{π} \int_{0}^{π} e^{u / π} sin u d u = \frac{π ( e + 1 )}{1 + π ^{2}}

The sum of the first

n + 1

regions (from

j = 0

n

) is given by the geometric series sum formula

S = S_{0} \frac{r ^{n + 1} - 1}{r - 1}

. Substituting our values, we arrive at the final result:

$S = \frac{π ( 1 + e ) ( e ^{n + 1} - 1 )}{( 1 + π ^{2} ) ( e - 1 )}$

Common Traps & FAQs

Wait, why do we use the absolute value

∣ x ∣

? Isn't the area just the integral of the function?

Great question! Remember, the integral

\int f (x) d x

gives the *signed* area. If the curve dips below the axis, the integral becomes negative. But 'area' in geometry is a physical quantity—it must be positive. Since our curve

x = e^{- a y} sin (b y)

oscillates, it crosses the

y

-axis. To get the total area, we must integrate the magnitude

∣ x ∣

Why does the number of terms in the sum become

n + 1

This is a classic trap! We are summing

S_{j}

for

j = 0, 1, 2, \dots, n

. If you count the integers from

0

n

, you have

n

numbers plus the zero itself, giving you

n + 1

terms. Always double-check your indices—it's the difference between a correct answer and a silly mistake.

Is there a trick to the integration of

e^{u / π} sin (u)

Yes! Don't waste time doing integration by parts twice. Memorize the standard form:

\int e^{A x} sin (B x) d x = \frac{e ^{A x}}{A ^{2} + B ^{2}} (A sin (B x) - B cos (B x))

. It is a lifesaver in JEE Advanced. Just identify

A

and

B

carefully, and the result falls out instantly.

Visualized Solution

Visualizing the Regions $S_{j}$

Setting up the Integral for $S_{j}$

Substitution $t = b y$

Shifting the Limits to $0$ to $π$

Simplifying the Integrand

Proving the Geometric Progression

Substituting $a = - 1$ and $b = π$

Integration Formula for $S_{0}$

Evaluating $S_{0}$

Sum of the Geometric Progression

Final Result

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Transformation

The Geometric Progression

Final Calculation

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Visualized Solution

Visualizing the Regions .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sj​

Setting up the Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sj​

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=by

Shifting the Limits to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0 to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } π

Simplifying the Integrand

Proving the Geometric Progression

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=−1 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=π

Integration Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S0​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S0​

Sum of the Geometric Progression

Final Result

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Transformation

The Geometric Progression

Final Calculation

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Integration Formula for $S_{0}$

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