JEE Main 2004

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: If $f (x) = \frac{e ^{x}}{1 + e ^{x}}$ , $I_{1} = \int_{f (- a)}^{f (a)} xg {x (1 - x)} d x$ and $I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x$ , then the value of $I_{2} / I_{1}$ is

Q: Why did the argument $x(1-x)$ stay the same?

When you substitute $x$ with $1-x$ in the expression $x(1-x)$, it becomes $(1-x)(1-(1-x))$, which simplifies to $(1-x)(x)$. It is a symmetric structure that is invariant under the transformation $x \to 1-x$. This is the 'magic' that makes the integral solvable.

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Visualized Solution

Problem Statement

Given function: $f (x) = \frac{e ^{x}}{1 + e ^{x}}$
Integral $I_{1} = \int_{f (- a)}^{f (a)} xg {x (1 - x)} d x$
Integral $I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x$
Objective: Find the value of $\frac{I _{2}}{I _{1}}$

Analyzing the Limits

Let's look at the limits of integration: $f (a)$ and $f (- a)$ .

Sum of Limits

Calculate the sum of the limits:
$f (a) + f (- a) = \frac{e ^{a}}{1 + e ^{a}} + \frac{e ^{- a}}{1 + e ^{- a}}$

Simplifying the Sum

Simplify the second term by multiplying numerator and denominator by $e^{a}$ :
$\frac{e ^{- a} \cdot e ^{a}}{( 1 + e ^{- a} ) \cdot e ^{a}} = \frac{1}{e ^{a} + 1}$

Final Sum of Limits

Add the simplified terms together:
$f (a) + f (- a) = \frac{e ^{a}}{1 + e ^{a}} + \frac{1}{e ^{a} + 1}$
$f (a) + f (- a) = \frac{e ^{a} + 1}{e ^{a} + 1} = 1$

Symmetry of Limits

Since $f (a) + f (- a) = 1$ , the limits are symmetric around their midpoint.
Midpoint $= \frac{f ( a ) + f ( - a )}{2} = 0.5$

King's Property of Integrals

Recall the King's Property of Definite Integrals:
$\int_{A}^{B} h (x) d x = \int_{A}^{B} h (A + B - x) d x$

Applying King's Property

For our integral $I_{1}$ , the sum of limits is $A + B = 1$ .
We will substitute $x \to 1 - x$ in $I_{1}$ .

Substituting $x \to 1 - x$

Apply the substitution to $I_{1}$ :
$I_{1} = \int_{f (- a)}^{f (a)} (1 - x) g {(1 - x) (1 - (1 - x))} d x$

Simplifying the Argument

Simplify the argument inside the function $g$ :
$(1 - x) (1 - (1 - x)) = (1 - x) (x) = x (1 - x)$

Rewriting $I_{1}$

Rewrite $I_{1}$ with the simplified argument:
$I_{1} = \int_{f (- a)}^{f (a)} (1 - x) g {x (1 - x)} d x$

Splitting the Integral

Expand the expression by splitting the integral:
$I_{1} = \int_{f (- a)}^{f (a)} 1 \cdot g {x (1 - x)} d x - \int_{f (- a)}^{f (a)} x \cdot g {x (1 - x)} d x$

Identifying Known Integrals

Identify the known integrals in the expanded form:
First part is $I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x$
Second part is $I_{1} = \int_{f (- a)}^{f (a)} xg {x (1 - x)} d x$

Forming the Equation

Substitute $I_{2}$ and $I_{1}$ back into the equation:
$I_{1} = I_{2} - I_{1}$

Calculating the Final Ratio

Solve for the ratio $\frac{I _{2}}{I _{1}}$ :
$2 I_{1} = I_{2}$
$\frac{I _{2}}{I _{1}} = 2$

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The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

We are given the function

f (x) = \frac{e ^{x}}{1 + e ^{x}}

and two integrals defined as:

I_{1} = \int_{f (- a)}^{f (a)} xg {x (1 - x)} d x

I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x

Our objective is to determine the ratio

\frac{I _{2}}{I _{1}}

The Mystery of the Limits

In problems involving definite integrals with complex limits, the first step is to examine the sum of the limits. Let us calculate

f (a) + f (- a)

Given

f (a) = \frac{e ^{a}}{1 + e ^{a}}

and

f (- a) = \frac{e ^{- a}}{1 + e ^{- a}}

, we simplify

f (- a)

by multiplying the numerator and denominator by

e^{a}

f (- a) = \frac{e ^{- a} \cdot e ^{a}}{( 1 + e ^{- a} ) \cdot e ^{a}} = \frac{1}{e ^{a} + 1}

Adding these together yields:

f (a) + f (- a) = \frac{e ^{a}}{1 + e ^{a}} + \frac{1}{1 + e ^{a}} = \frac{e ^{a} + 1}{1 + e ^{a}} = 1

This result is the key to the problem, as it reveals that the limits are symmetric around the value

0.5

The King's Property

Since the sum of the limits is

1

, we apply the King's Property of definite integrals, which states that

\int_{A}^{B} h (x) d x = \int_{A}^{B} h (A + B - x) d x

. Here,

A + B = 1

, so we substitute

x

with

(1 - x)

Applying this to

I_{1}

I_{1} = \int_{f (- a)}^{f (a)} (1 - x) g {(1 - x) (1 - (1 - x))} d x

Simplifying the argument inside the function

g

(1 - x) (1 - 1 + x) = (1 - x) (x) = x (1 - x)

The function

g

remains invariant under this transformation.

The Elegant Cancellation

Now, we expand the expression for

I_{1}

using the property derived above:

I_{1} = \int_{f (- a)}^{f (a)} 1 \cdot g {x (1 - x)} d x - \int_{f (- a)}^{f (a)} x \cdot g {x (1 - x)} d x

Observe that the first integral is exactly

I_{2}

, and the second integral is our original

I_{1}

. We can write this as:

I_{1} = I_{2} - I_{1}

Rearranging the terms, we obtain:

2 I_{1} = I_{2}

Therefore, the final ratio is:

\frac{I _{2}}{I _{1}} = 2

Common Traps & FAQs

Why does

f (a) + f (- a) = 1

It is a beautiful property of the logistic function. When you add

f (x) = \frac{e ^{x}}{1 + e ^{x}}

and

f (- x) = \frac{e ^{- x}}{1 + e ^{- x}}

, the second term simplifies to

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

by multiplying the numerator and denominator by

e^{x}

. Adding them gives

\frac{e ^{x} + 1}{1 + e ^{x}} = 1

. It is a classic trick to simplify symmetric limits!

Can we always use the King's Property?

The King's Property is a powerful tool, but it requires the limits to be constants or symmetric. Here, the sum of the limits is a constant, which is the perfect condition for its application. Always check the sum of your limits first!

Why did the argument

x (1 - x)

stay the same?

When you substitute

x

with

1 - x

in the expression

x (1 - x)

, it becomes

(1 - x) (1 - (1 - x))

, which simplifies to

(1 - x) (x)

. It is a symmetric structure that is invariant under the transformation

x \to 1 - x

. This is the 'magic' that makes the integral solvable.

Animated Solution for Mathematics - Definite Integration: If $f (x) = \frac{e ^{x}}{1 + e ^{x}}$ , $I_{1} = \int_{f (- a)}^{f (a)} xg {x (1 - x)} d x$ and $I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x$ , then the value of $I_{2} / I_{1}$ is

Select Answer:

Visualized Solution

Problem Statement

Analyzing the Limits

Sum of Limits

Simplifying the Sum

Final Sum of Limits

Symmetry of Limits

King's Property of Integrals

Applying King's Property

Substituting $x \to 1 - x$

Simplifying the Argument

Rewriting $I_{1}$

Splitting the Integral

Identifying Known Integrals

Forming the Equation

Calculating the Final Ratio

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Mystery of the Limits

The King's Property

The Elegant Cancellation

Similar Questions

Let $f (x)$ be a positive function and $I_{1} = \int_{1/2}^{1} 2 x f (2 x (1 - 2 x)) d x$ and $I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I _{2}}{I _{1}}$ is equal to ________

The integral $\int_{1}^{e} {(\frac{x}{e})^{2 x} - (\frac{e}{x})^{x}} lo g_{e} x d x$ is equal to :

The value of the integral $\int_{e^{- 1}}^{e^{2}} ∣ \frac{l o g _{e} x}{x} ∣ d x$ is

If $f (x) = \frac{2 - x c o s x}{2 + x c o s x}$ and $g (x) = lo g_{e} x, (x > 0)$ then the value of integral $\int_{- π /4}^{π /4} g (f (x)) d x$ is :

The value of $\int_{- 1}^{1} \frac{( 1 + ∣ x ∣ - x ) e ^{x} + ( ∣ x ∣ - x ) e ^{- x}}{e ^{x} + e ^{- x}} d x$ is equal to

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e ^{((l o g_{e} x)^{2} + 1)^{- 1}}}{e ^{((l o g_{e} x)^{2} + 1)^{- 1}} + e ^{((6 - l o g_{e} x)^{2} + 1)^{- 1}}}) d x$ is

The value of $\int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$ is equal to

Let the function $f : [0, 2] \to R$ be defined as $f (x) = {e^{m i n {x^{2}, x - [x]}}, e^{[x - l n x]}, x \in [0, 1) x \in [1, 2]$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is

The value of the integral $\int_{- 1}^{1} lo g_{e} (1 - x + 1 + x) d x$ is equal to :

The value of the integral $\int_{- 2}^{2} \frac{∣ x ^{3} + x ∣}{( e ^{x ∣ x ∣} + 1 )} d x$ is equal to

Select Answer:

Visualized Solution

Problem Statement

Analyzing the Limits

Sum of Limits

Simplifying the Sum

Final Sum of Limits

Symmetry of Limits

King's Property of Integrals

Applying King's Property

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→1−x

Simplifying the Argument

Rewriting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Splitting the Integral

Identifying Known Integrals

Forming the Equation

Calculating the Final Ratio

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Mystery of the Limits

The King's Property

The Elegant Cancellation

Similar Questions

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Substituting $x \to 1 - x$

Rewriting $I_{1}$

Let $f (x)$ be a positive function and $I_{1} = \int_{1/2}^{1} 2 x f (2 x (1 - 2 x)) d x$ and $I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I _{2}}{I _{1}}$ is equal to ________

The integral $\int_{1}^{e} {(\frac{x}{e})^{2 x} - (\frac{e}{x})^{x}} lo g_{e} x d x$ is equal to :

The value of the integral $\int_{e^{- 1}}^{e^{2}} ∣ \frac{l o g _{e} x}{x} ∣ d x$ is

If $f (x) = \frac{2 - x c o s x}{2 + x c o s x}$ and $g (x) = lo g_{e} x, (x > 0)$ then the value of integral $\int_{- π /4}^{π /4} g (f (x)) d x$ is :

The value of $\int_{- 1}^{1} \frac{( 1 + ∣ x ∣ - x ) e ^{x} + ( ∣ x ∣ - x ) e ^{- x}}{e ^{x} + e ^{- x}} d x$ is equal to

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e ^{((l o g_{e} x)^{2} + 1)^{- 1}}}{e ^{((l o g_{e} x)^{2} + 1)^{- 1}} + e ^{((6 - l o g_{e} x)^{2} + 1)^{- 1}}}) d x$ is

The value of $\int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$ is equal to

The value of the integral $\int_{- 1}^{1} lo g_{e} (1 - x + 1 + x) d x$ is equal to :

The value of the integral $\int_{- 2}^{2} \frac{∣ x ^{3} + x ∣}{( e ^{x ∣ x ∣} + 1 )} d x$ is equal to