JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{∣ x ∣} - e^{- x}, a > 0}$ is $\frac{e ^{2} + 8 e + 1}{e}$ then the value of a is :

Q: Wait, why did the $e^{-x}$ term not simplify using symmetry?

Great observation! The function $e^{-x}$ is neither even nor odd. If you replace $x$ with $-x$, you get $e^{x}$, which is not the same as $e^{-x}$ (even) or $-e^{-x}$ (odd). Therefore, it does not possess the symmetry required to use the $2 \int_{0}^{a}$ trick. You must integrate it directly.

Q: I got lost in the signs when integrating $e^{-x}$. Any tips?

The integral of $e^{-x}$ is $-e^{-x}$. When evaluating from $-1$ to $1$, you get $[-e^{-x}]_{-1}^{1} = (-e^{-1}) - (-e^{-1 \cdot -1}) = -\frac{1}{e} - (-e) = e - \frac{1}{e}$. Always use parentheses to keep those negative signs from swallowing your answer!

Select Answer:

Progressive Hints

Hint 1

The Spark: Look closely at the interval of integration,

[- 1, 1]

. It is perfectly symmetric about the origin. Now, examine the function

f (x) = a + e^{∣ x ∣} - e^{- x}

. Does the term

e^{∣ x ∣}

behave differently than

e^{- x}

? Recognizing the even symmetry of the absolute value function is your first step to victory.

Hint 2

The Arsenal: You need the fundamental property of definite integrals for even functions:

\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x

f (x)

is even. Apply this specifically to the

e^{∣ x ∣}

term. For the other terms, use the standard fundamental theorem of calculus.

Hint 3

The Execution: Break the integral into three distinct parts:

A = \int_{- 1}^{1} a d x + \int_{- 1}^{1} e^{∣ x ∣} d x - \int_{- 1}^{1} e^{- x} d x

. Evaluate the first part as a rectangle, the second using the symmetry property, and the third using direct integration. Set the sum equal to

\frac{e ^{2} + 8 e + 1}{e}

and solve for

a

Visualized Solution

Visualizing the Region

Region: ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{∣ x ∣} - e^{- x}}$
Lower bound: $y = 0$ (x-axis)
Upper bound: $y = a + e^{∣ x ∣} - e^{- x}$

The Integral Setup

Area $A = \int_{- 1}^{1} (a + e^{∣ x ∣} - e^{- x}) d x$
Given Area $= \frac{e ^{2} + 8 e + 1}{e}$

Splitting the Integral

$A = \int_{- 1}^{1} a d x + \int_{- 1}^{1} e^{∣ x ∣} d x - \int_{- 1}^{1} e^{- x} d x$

Integrating the Constant

$\int_{- 1}^{1} a d x = [a x]_{- 1}^{1}$
$= a (1 - (- 1)) = 2 a$

Handling the Absolute Value

$e^{∣ x ∣}$ is an even function
$\int_{- 1}^{1} e^{∣ x ∣} d x = 2 \int_{0}^{1} e^{x} d x$

Evaluating Part 2

$2 [e^{x}]_{0}^{1} = 2 (e^{1} - e^{0})$
$= 2 (e - 1)$

Integrating the Exponential

$\int_{- 1}^{1} e^{- x} d x = [- e^{- x}]_{- 1}^{1}$
$= - e^{- 1} - (- e^{1}) = e - \frac{1}{e}$

Summing the Components

$A = 2 a + 2 (e - 1) - (e - \frac{1}{e})$
$A = 2 a + 2 e - 2 - e + \frac{1}{e}$

Algebraic Simplification

$A = 2 a - 2 + e + \frac{1}{e}$
$A = 2 a - 2 + \frac{e ^{2} + 1}{e}$

Equating to the Given Area

$2 a - 2 + \frac{e ^{2} + 1}{e} = \frac{e ^{2} + 8 e + 1}{e}$
$2 a - 2 + \frac{e ^{2} + 1}{e} = \frac{e ^{2} + 1}{e} + \frac{8 e}{e}$
$2 a - 2 + \frac{e ^{2} + 1}{e} = \frac{e ^{2} + 1}{e} + 8$

Finding the Value of $a$

$2 a - 2 = 8$
$2 a = 10$
$a = 5$

The Way Forward

Key Takeaway: Use symmetry (even/odd) to simplify integrals involving $∣ x ∣$ .
Final Answer: $a = 5$
Next Challenge: What if the region was bounded by $y = a - e^{∣ x ∣}$ ? How would the area change?

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

We are tasked with finding the constant

a

such that the area under the curve

y = a + e^{∣ x ∣} - e^{- x}

over the interval

x \in [- 1, 1]

equals

\frac{e ^{2} + 8 e + 1}{e}

The total area

A

is defined by the definite integral:

A = \int_{- 1}^{1} (a + e^{∣ x ∣} - e^{- x}) d x

By the linearity of the integral, we can decompose this into three distinct parts:

A = \int_{- 1}^{1} a d x + \int_{- 1}^{1} e^{∣ x ∣} d x - \int_{- 1}^{1} e^{- x} d x

Evaluating the Components

The first integral represents the area of a rectangle with width

2

and height

a

\int_{- 1}^{1} a d x = 2 a

For the second integral, we utilize the property that

e^{∣ x ∣}

is an even function. Thus, we can simplify the integration range:

\int_{- 1}^{1} e^{∣ x ∣} d x = 2 \int_{0}^{1} e^{x} d x = 2 [e^{x}]_{0}^{1} = 2 (e - 1)

For the third integral, we evaluate the antiderivative of

e^{- x}

\int_{- 1}^{1} e^{- x} d x = [- e^{- x}]_{- 1}^{1} = (- e^{- 1}) - (- e^{1}) = e - \frac{1}{e}

The Master Equation

Combining these results, the total area

A

is:

A = 2 a + 2 (e - 1) - (e - \frac{1}{e})

Expanding the expression, we obtain:

A = 2 a + 2 e - 2 - e + \frac{1}{e} = 2 a - 2 + e + \frac{1}{e}

We can rewrite the expression

e + \frac{1}{e}

\frac{e ^{2} + 1}{e}

A = 2 a - 2 + \frac{e ^{2} + 1}{e}

Final Calculation

We set our expression for

A

equal to the target value provided in the problem:

2 a - 2 + \frac{e ^{2} + 1}{e} = \frac{e ^{2} + 8 e + 1}{e}

The right side can be decomposed as:

\frac{e ^{2} + 1}{e} + \frac{8 e}{e} = \frac{e ^{2} + 1}{e} + 8

Subtracting

\frac{e ^{2} + 1}{e}

from both sides, we are left with:

2 a - 2 = 8

Solving for

a

2 a = 10 \Rightarrow * * a = 5 * *

Common Traps & FAQs

Why can't we just integrate

e^{∣ x ∣}

directly without splitting it?

You technically could, but it would require splitting the integral at

x = 0

because the definition of

∣ x ∣

changes at the origin. By recognizing that

e^{∣ x ∣}

is an even function, you can use the symmetry property

\int_{- 1}^{1} e^{∣ x ∣} d x = 2 \int_{0}^{1} e^{x} d x

, which is much faster and less prone to sign errors.

Wait, why did the

e^{- x}

term not simplify using symmetry?

Great observation! The function

e^{- x}

is neither even nor odd. If you replace

x

with

- x

, you get

e^{x}

, which is not the same as

e^{- x}

(even) or

- e^{- x}

(odd). Therefore, it does not possess the symmetry required to use the

2 \int_{0}^{a}

trick. You must integrate it directly.

I got lost in the signs when integrating

e^{- x}

. Any tips?

The integral of

e^{- x}

- e^{- x}

. When evaluating from

- 1

1

, you get

[- e^{- x}]_{- 1}^{1} = (- e^{- 1}) - (- e^{- 1 \cdot- 1}) = - \frac{1}{e} - (- e) = e - \frac{1}{e}

. Always use parentheses to keep those negative signs from swallowing your answer!

Animated Solution for Mathematics - Definite Integration: If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{∣ x ∣} - e^{- x}, a > 0}$ is $\frac{e ^{2} + 8 e + 1}{e}$ then the value of a is :

Select Answer:

Visualized Solution

Visualizing the Region

The Integral Setup

Splitting the Integral

Integrating the Constant

Handling the Absolute Value

Evaluating Part 2

Integrating the Exponential

Summing the Components

Algebraic Simplification

Equating to the Given Area

Finding the Value of $a$

The Way Forward

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Evaluating the Components

The Master Equation

Final Calculation

Similar Questions

If the area of the region ${(x, y) : \frac{a}{x ^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(lo g_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to:

If the area of the bounded region $R = {(x, y) : max {0, lo g_{e} x} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2}$ is, $α (lo g_{e} 2)^{- 1} + β (lo g_{e} 2) + γ$ , then the value of $(α + β - 2 γ)^{2}$ is equal to :

The area of the region ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is

The area of the region $R = {(x, y) : x y \leq 8, 1 \leq y \leq x^{2}, x \geq 0}$ is

The area enclosed between the curves $y = a x^{2}$ and $x = a y^{2}$ ( $a > 0$ ) is 1 sq. unit, then the value of $a$ is

Find the area bounded by the x-axis, part of the curve $y = (1 + \frac{8}{x ^{2}})$ and the ordinates at $x = 2$ and $x = 4$ . If the ordinate at $x = a$ divides the area into two equal parts, find $a$ .

If the area (in sq. units) of the region ${(x, y) : y^{2} \leq 4 x, x + y \leq 1, x \geq 0, y \geq 0}$ is $a 2 + b$ , then $a - b$ is equal to :

Let $α$ be the area of the larger region bounded by the curve $y^{2} = 8 x$ and the lines $y = x$ and $x = 2$ , which lies in the first quadrant. Then the value of $3 α$ is equal to ______.

Let the area of the region ${(x, y) : 2 y \leq x^{2} + 3, y + ∣ x ∣ \leq 3, y \geq ∣ x - 1∣}$ be A. Then 6A is equal to :

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is

Select Answer:

Visualized Solution

Visualizing the Region

The Integral Setup

Splitting the Integral

Integrating the Constant

Handling the Absolute Value

Evaluating Part 2

Integrating the Exponential

Summing the Components

Algebraic Simplification

Equating to the Given Area

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

The Way Forward

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Evaluating the Components

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the area of the region {(x,y):x2a​≤y≤x1​,1≤x≤2,0<a<1} is (loge​2)−71​ then the value of 7a−3 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the area of the bounded region R={(x,y):max{0,loge​x}≤y≤2x,21​≤x≤2} is, α(loge​2)−1+β(loge​2)+γ, then the value of (α+β−2γ)2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region {(x,y):xy≤8,1≤y≤x2} is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region R={(x,y):xy≤8,1≤y≤x2,x≥0} is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area enclosed between the curves y=ax2 and x=ay2 (a>0) is 1 sq. unit, then the value of a is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the area bounded by the x-axis, part of the curve y=(1+x28​) and the ordinates at x=2 and x=4. If the ordinate at x=a divides the area into two equal parts, find a.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the area (in sq. units) of the region {(x,y):y2≤4x,x+y≤1,x≥0,y≥0} is a2​+b, then a−b is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α be the area of the larger region bounded by the curve y2=8x and the lines y=x and x=2, which lies in the first quadrant. Then the value of 3α is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the area of the region {(x,y):2y≤x2+3,y+∣x∣≤3,y≥∣x−1∣} be A. Then 6A is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region {(x,y):x2≤y≤8−x2,y≤7} is

Finding the Value of $a$

If the area of the region ${(x, y) : \frac{a}{x ^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(lo g_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to:

If the area of the bounded region $R = {(x, y) : max {0, lo g_{e} x} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2}$ is, $α (lo g_{e} 2)^{- 1} + β (lo g_{e} 2) + γ$ , then the value of $(α + β - 2 γ)^{2}$ is equal to :

The area of the region ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is

The area of the region $R = {(x, y) : x y \leq 8, 1 \leq y \leq x^{2}, x \geq 0}$ is

The area enclosed between the curves $y = a x^{2}$ and $x = a y^{2}$ ( $a > 0$ ) is 1 sq. unit, then the value of $a$ is

Find the area bounded by the x-axis, part of the curve $y = (1 + \frac{8}{x ^{2}})$ and the ordinates at $x = 2$ and $x = 4$ . If the ordinate at $x = a$ divides the area into two equal parts, find $a$ .

If the area (in sq. units) of the region ${(x, y) : y^{2} \leq 4 x, x + y \leq 1, x \geq 0, y \geq 0}$ is $a 2 + b$ , then $a - b$ is equal to :

Let $α$ be the area of the larger region bounded by the curve $y^{2} = 8 x$ and the lines $y = x$ and $x = 2$ , which lies in the first quadrant. Then the value of $3 α$ is equal to ______.

Let the area of the region ${(x, y) : 2 y \leq x^{2} + 3, y + ∣ x ∣ \leq 3, y \geq ∣ x - 1∣}$ be A. Then 6A is equal to :

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is