JEE Main 2025 April

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $f : [0, \infty) \to R$ be differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then the area of the region bounded by $y = f (x)$ and the coordinate axes is

Q: Wait, why can't we just integrate the integral directly? It looks like a standard integral.

That is a common trap! The integral is a function of $x$ because the upper limit is $x$. You cannot treat it as a constant. You must differentiate to convert the integral equation into a differential equation. The integral is not a static value; it is a dynamic process that evolves with $x$.

Q: Why did we use $e^{-2x}$ as the integrating factor? How do I know it's not $e^{2x}$?

Great question. The integrating factor is defined as $e^{\int P(x) dx}$. In our derived equation $f'(x) - 2f(x) = 2x - 3$, the coefficient $P(x)$ is $-2$. Thus, the integral is $\int -2 dx = -2x$. Hence, $e^{-2x}$ is the correct factor. Always check the sign of $P(x)$ carefully!

Q: How did we get $f(0) = 1$ so easily?

Look at the original equation: $f(x) = 1 - 2x + \int_0^x e^{x-t}f(t)dt$. If you substitute $x=0$, the integral becomes $\int_0^0 ... dt$, which is identically zero. The term $-2x$ also becomes zero. You are left with $f(0) = 1 - 0 + 0 = 1$. It is a gift from the problem setter to help you find the constant of integration!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look closely at the structure of the equation

f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t

. The variable

x

is trapped both inside the integral limit and within the integrand itself. To solve this, you must liberate

f (x)

by differentiating with respect to

x

. Think of the integral as a dynamic accumulator that changes as

x

grows.

Hint 2

Hint 2: The Arsenal. You need the Leibniz Rule for differentiation under the integral sign. Remember, when you differentiate

\int_{0}^{x} g (x, t) d t

, you must account for both the explicit

x

in the integrand and the

x

in the upper limit. Also, recognize that

e^{x - t} = e^{x} e^{- t}

allows you to treat

e^{x}

as a constant relative to

t

, which is the key to simplifying the expression.

Hint 3

Hint 3: The Execution. Start by differentiating both sides:

f^{'} (x) = - 2 + \frac{d}{d x} (e^{x} \int_{0}^{x} e^{- t} f (t) d t)

. Apply the product rule to the term on the right, and use the Leibniz rule to handle the derivative of the integral. This will lead you to a linear differential equation of the form

f^{'} (x) + P (x) f (x) = Q (x)

Visualized Solution

Understanding the Integral Equation for $f (x)$

Given equation: $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$
This is an Integral Equation where the unknown function $f (x)$ appears both inside and outside the integral.
To solve for $f (x)$ , we need to convert this into a Differential Equation.

Separating the Variables $x$ and $t$

Rewrite the integral using exponent properties: $e^{x - t} = e^{x} \cdot e^{- t}$
Since the integral is with respect to $t$ , $e^{x}$ acts as a constant and can be pulled out:
$f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

Applying the Leibniz Rule for $\frac{d}{d x}$

Differentiating both sides with respect to $x$ using the Product Rule and Leibniz Rule:
$f^{'} (x) = \frac{d}{d x} (1) - \frac{d}{d x} (2 x) + \frac{d}{d x} (e^{x} \cdot \int_{0}^{x} e^{- t} f (t) d t)$
$f^{'} (x) = 0 - 2 + [e^{x} \int_{0}^{x} e^{- t} f (t) d t + e^{x} \cdot \frac{d}{d x} (\int_{0}^{x} e^{- t} f (t) d t)]$

Simplifying the Derivative $f^{'} (x)$

Applying Leibniz Rule: $\frac{d}{d x} \int_{0}^{x} e^{- t} f (t) d t = e^{- x} f (x)$
Substituting back: $f^{'} (x) = - 2 + e^{x} \int_{0}^{x} e^{- t} f (t) d t + e^{x} (e^{- x} f (x))$
$f^{'} (x) = - 2 + e^{x} \int_{0}^{x} e^{- t} f (t) d t + f (x)$

Forming the Linear Differential Equation $\frac{d y}{d x} + P y = Q$

From the original equation: $e^{x} \int_{0}^{x} e^{- t} f (t) d t = f (x) - 1 + 2 x$
Substitute this into the derivative expression:
$f^{'} (x) = - 2 + [f (x) - 1 + 2 x] + f (x)$
$f^{'} (x) - 2 f (x) = 2 x - 3$
This is a Linear Differential Equation of the form $\frac{d y}{d x} + P y = Q$ .

Finding the Integrating Factor $e^{\int P d x}$

Identify $P = - 2$ and $Q = 2 x - 3$ .
Integrating Factor (I.F.) $= e^{\int P d x} = e^{\int - 2 d x} = e^{- 2 x}$
The solution is given by: $f (x) \cdot I.F. = \int Q \cdot I.F. d x$
$f (x) \cdot e^{- 2 x} = \int (2 x - 3) e^{- 2 x} d x$

Integration by Parts $\int u d v$

Using Integration by Parts: $\int u d v = uv - \int v d u$
Let $u = 2 x - 3 \Rightarrow d u = 2 d x$
Let $d v = e^{- 2 x} d x \Rightarrow v = - \frac{1}{2} e^{- 2 x}$
$\int (2 x - 3) e^{- 2 x} d x = (2 x - 3) (- \frac{1}{2} e^{- 2 x}) - \int (- \frac{1}{2} e^{- 2 x}) (2) d x$

Simplifying the General Solution for $f (x)$

$f (x) e^{- 2 x} = - \frac{2 x - 3}{2} e^{- 2 x} + \int e^{- 2 x} d x$
$f (x) e^{- 2 x} = - \frac{2 x - 3}{2} e^{- 2 x} - \frac{1}{2} e^{- 2 x} + c$
Multiply by $e^{2 x}$ to isolate $f (x)$ :
$f (x) = - \frac{2 x - 3}{2} - \frac{1}{2} + c e^{2 x}$
$f (x) = - x + \frac{3}{2} - \frac{1}{2} + c e^{2 x} = 1 - x + c e^{2 x}$

Finding the Constant of Integration $c$

Find $f (0)$ from the original equation: $f (0) = 1 - 2 (0) + \int_{0}^{0} e^{0 - t} f (t) d t = 1$
Substitute $x = 0, f (0) = 1$ into $f (x) = 1 - x + c e^{2 x}$ :
$1 = 1 - 0 + c \cdot e^{0} \Rightarrow 1 = 1 + c \Rightarrow c = 0$
The function is exactly: $f (x) = 1 - x$

Visualizing the Region for $y = 1 - x$

The function is $y = 1 - x$ .
Intercepts with axes: $(1, 0)$ and $(0, 1)$ .
The region is a triangle in the first quadrant bounded by $x = 0, y = 0$ , and $y = 1 - x$ .

Final Area Calculation

Area of the triangular region $= \frac{1}{2} \times base \times height$
Base $= 1$ unit, Height $= 1$ unit.
Area $= \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$
Final Answer: The area is $\frac{1}{2}$ square units.

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Welcome, future engineers. Today, we are going to dismantle a problem that looks intimidating at first glance but reveals a beautiful, simple truth once we peel back the layers.

We are dealing with an integral equation:

f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t

Many students freeze when they see an unknown function

f (x)

trapped inside an integral. But I want you to see this not as a cage, but as a challenge. Our goal is to liberate

f (x)

by transforming this integral equation into a differential equation.

The Transformation Phase

First, let us look at the term inside the integral:

e^{x - t}

. Since the integral is with respect to

t

, the term

e^{x}

is effectively a constant.

We can pull it out:

f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t

Now, the structure is much clearer. We have a product of

e^{x}

and an integral that depends on

x

. This is the perfect setup for the Leibniz Rule.

The Calculus of Liberation

We differentiate both sides with respect to

x

. On the left, we get

f^{'} (x)

. On the right, the derivative of

1

0

, and the derivative of

- 2 x

- 2

For the integral term, we must use the product rule:

\frac{d}{d x} (e^{x} \int_{0}^{x} e^{- t} f (t) d t) = e^{x} \int_{0}^{x} e^{- t} f (t) d t + e^{x} \cdot \frac{d}{d x} (\int_{0}^{x} e^{- t} f (t) d t)

Applying the Leibniz Rule to the derivative of the integral, we get

e^{- x} f (x)

. Notice the magic:

e^{x} \cdot e^{- x} = 1

. The exponential terms vanish!

We are left with:

f^{'} (x) = - 2 + e^{x} \int_{0}^{x} e^{- t} f (t) d t + f (x)

The Linear ODE

We still have that integral term. But look back at our original equation! We can rearrange it to find that:

e^{x} \int_{0}^{x} e^{- t} f (t) d t = f (x) - 1 + 2 x

Substituting this into our derivative equation, we get:

f^{'} (x) = - 2 + [f (x) - 1 + 2 x] + f (x)

Simplifying this, we arrive at:

f^{'} (x) - 2 f (x) = 2 x - 3

This is a standard linear differential equation of the form

\frac{d y}{d x} + P y = Q

. Our integrating factor is

e^{\int - 2 d x} = e^{- 2 x}

Multiplying through, we get:

f (x) e^{- 2 x} = \int (2 x - 3) e^{- 2 x} d x

Using integration by parts, we solve the right side to find:

f (x) = 1 - x + C e^{2 x}

The Final Reveal

We use the initial condition

f (0) = 1

to find that

C = 0

. Our function is simply

f (x) = 1 - x

This is a straight line! The area bounded by

y = 1 - x

and the coordinate axes is a simple right-angled triangle with base

1

and height

1

The area is:

\frac{1}{2} \times 1 \times 1 = \frac{1}{2}

We have turned a complex integral equation into a simple geometric shape. That is the power of calculus. The final answer is $1/2$ .

Common Traps & FAQs

Wait, why can't we just integrate the integral directly? It looks like a standard integral.

That is a common trap! The integral is a function of

x

because the upper limit is

x

. You cannot treat it as a constant. You must differentiate to convert the integral equation into a differential equation. The integral is not a static value; it is a dynamic process that evolves with

x

Why did we use

e^{- 2 x}

as the integrating factor? How do I know it's not

e^{2 x}

Great question. The integrating factor is defined as

e^{\int P (x) d x}

. In our derived equation

f^{'} (x) - 2 f (x) = 2 x - 3

, the coefficient

P (x)

- 2

. Thus, the integral is

\int - 2 d x = - 2 x

. Hence,

e^{- 2 x}

is the correct factor. Always check the sign of

P (x)

carefully!

How did we get

f (0) = 1

so easily?

Look at the original equation:

f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t

. If you substitute

x = 0

, the integral becomes

\int_{0}^{0} ... d t

, which is identically zero. The term

- 2 x

also becomes zero. You are left with

f (0) = 1 - 0 + 0 = 1

. It is a gift from the problem setter to help you find the constant of integration!

Animated Solution for Mathematics - Differential Equations: Let $f : [0, \infty) \to R$ be differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then the area of the region bounded by $y = f (x)$ and the coordinate axes is

Select Answer:

Visualized Solution

Understanding the Integral Equation for $f (x)$

Separating the Variables $x$ and $t$

Applying the Leibniz Rule for $\frac{d}{d x}$

Simplifying the Derivative $f^{'} (x)$

Forming the Linear Differential Equation $\frac{d y}{d x} + P y = Q$

Finding the Integrating Factor $e^{\int P d x}$

Integration by Parts $\int u d v$

Simplifying the General Solution for $f (x)$

Finding the Constant of Integration $c$

Visualizing the Region for $y = 1 - x$

Final Area Calculation

The Sigma Insight: Linear Differential Equations

The Transformation Phase

The Calculus of Liberation

The Linear ODE

The Final Reveal

Similar Questions

Let $f : [0, \infty) \to R$ be a continuous function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then, which of the following statement(s) is (are) TRUE ?

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{( 1 + x ^{2} ) ^{2}} y = x e^{- \frac{1}{( 1 + x ^{2} )}}; y (0) = 0$ . Then the area enclosed by the curve $f (x) = y (x) e^{\frac{1}{( 1 + x ^{2} )}}$ and the line $y - x = 4$ is__________

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ If $f (x) = 1$ , then $x$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

Let $f : R \to R$ be a differentiable function such that $f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t$ . If $f (0) = e^{- 2}$ , then $2 f (0) - f (2)$ is equal to ______.

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x$ is equal to

Select Answer:

Visualized Solution

Understanding the Integral Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Separating the Variables .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Applying the Leibniz Rule for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxd​

Simplifying the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Forming the Linear Differential Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​+Py=Q

Finding the Integrating Factor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e∫Pdx

Integration by Parts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫udv

Simplifying the General Solution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

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

Visualizing the Region for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=1−x

Final Area Calculation

The Sigma Insight: Linear Differential Equations

The Transformation Phase

The Calculus of Liberation

The Linear ODE

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[0,∞)→R be a continuous function such that f(x)=1−2x+∫0x​ex−tf(t)dt for all x∈[0,∞). Then, which of the following statement(s) is (are) TRUE ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function y=f(x) is the solution of the differential equation dxdy​+x2−1xy​=1−x2​x4+2x​ in (−1,1) satisfying f(0)=0. Then ∫−3​/23​/2​f(x)dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,∞)→(0,∞) be a differentiable function such that f(1)=e and limt→x​t−xt2f2(x)−x2f2(t)​=0 If f(x)=1, then x is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​=(y+1)((y+1)ex2/2−x),0<x<2.1, with y(2)=0. Then the value of dxdy​ at x=1 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function such that f′(x)+f(x)=∫02​f(t)dt. If f(0)=e−2, then 2f(0)−f(2) is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[1,∞)→[2,∞) be a differentiable function such that f(1)=2. If 6∫1x​f(t)dt=3xf(x)−x3 for all x≥1, then the value of f(2) is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be differentiable on the interval (0,∞) such that f(1)=1, and limt→x​t−xt2f(x)−x2f(t)​=1 for each x>0. Then f(x) is

Understanding the Integral Equation for $f (x)$

Separating the Variables $x$ and $t$

Applying the Leibniz Rule for $\frac{d}{d x}$

Simplifying the Derivative $f^{'} (x)$

Forming the Linear Differential Equation $\frac{d y}{d x} + P y = Q$

Finding the Integrating Factor $e^{\int P d x}$

Integration by Parts $\int u d v$

Simplifying the General Solution for $f (x)$

Finding the Constant of Integration $c$

Visualizing the Region for $y = 1 - x$

Let $f : [0, \infty) \to R$ be a continuous function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then, which of the following statement(s) is (are) TRUE ?

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ If $f (x) = 1$ , then $x$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

Let $f : R \to R$ be a differentiable function such that $f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t$ . If $f (0) = e^{- 2}$ , then $2 f (0) - f (2)$ is equal to ______.

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is