JEE Main 2021 (17 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be defined as $f (x) = e^{- x} sin x$ . If $F : [0, 1] \to R$ is a differentiable function such that $F (x) = \int_{0}^{x} f (t) d t$ , then the value of $\int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x$ lies in the interval

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Visualized Solution

Defining the Functions

Given function: $f (x) = e^{- x} sin x$
Integral function: $F (x) = \int_{0}^{x} f (t) d t$
Target: Evaluate $\int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x$

Fundamental Theorem of Calculus

To find $F^{'} (x)$ , we use the First Fundamental Theorem of Calculus (Leibniz Rule).
$\frac{d}{d x} \int_{0}^{x} f (t) d t = f (x)$

Differentiating $F (x)$

Applying the theorem: $F^{'} (x) = f (x)$
Therefore, $F^{'} (x) = e^{- x} sin x$

Substituting into the Target Integral

Let $I = \int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x$
Substitute $F^{'} (x) = f (x)$ into the expression.
$I = \int_{0}^{1} (f (x) + f (x)) e^{x} d x$

Simplifying the Integrand

Add the terms inside the bracket: $f (x) + f (x) = 2 f (x)$
The integral becomes: $I = \int_{0}^{1} 2 f (x) e^{x} d x$

Substituting $f (x)$

Recall the original definition: $f (x) = e^{- x} sin x$
Substitute this back into the integral.
$I = \int_{0}^{1} 2 (e^{- x} sin x) e^{x} d x$

Canceling Exponential Terms

Notice the product: $e^{- x} \cdot e^{x} = e^{0} = 1$
The integrand simplifies drastically.
$I = \int_{0}^{1} 2 sin x d x$

Visualizing the Definite Integral

The integral represents the area under $y = 2 sin x$ from $x = 0$ to $x = 1$ .
We can pull the constant $2$ outside: $I = 2 \int_{0}^{1} sin x d x$

Integrating Sine Function

The anti-derivative of $sin x$ is $- cos x$ .
$I = 2 [- cos x]_{0}^{1}$

Applying the Limits

Substitute upper limit $1$ and lower limit $0$ .
$I = 2 (- cos 1 - (- cos 0))$
Since $cos 0 = 1$ , this becomes:
$I = 2 (1 - cos 1)$

Taylor Series Expansion

We need the numerical value of $cos 1$ .
Use the Maclaurin/Taylor series for $cos x$ :
$cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \frac{x ^{6}}{6 !} + \dots$

Expanding $cos 1$

Substitute $x = 1$ into the series.
$cos 1 = 1 - \frac{1 ^{2}}{2 !} + \frac{1 ^{4}}{4 !} - \frac{1 ^{6}}{6 !} + \dots$
$cos 1 = 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} + \dots$

Substituting Series into Integral

Recall $I = 2 (1 - cos 1)$
Substitute the series for $cos 1$ :
$I = 2 (1 - (1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} + \dots))$

Simplifying the Series

The $1$ and $- 1$ cancel out.
Distribute the negative sign:
$I = 2 (\frac{1}{2} - \frac{1}{24} + \frac{1}{720} - \dots)$
Multiply by $2$ :
$I = 1 - \frac{1}{12} + \frac{1}{360} - \dots$

Bounding the Alternating Series

Calculate the sum of the first three terms:
$1 - \frac{1}{12} + \frac{1}{360} = \frac{360 - 30 + 1}{360} = \frac{331}{360}$
The series is alternating and terms are decreasing.
The next term is negative, so $I < \frac{331}{360}$ .
The previous partial sum was $1 - \frac{1}{12} = \frac{330}{360}$ , so $I > \frac{330}{360}$ .
Therefore, $I \in [\frac{330}{360}, \frac{331}{360}]$ .

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The Sigma Insight: Newton-Leibniz & Reduction Formulas

Solution Diagram

Analyzing the Setup

Welcome, future engineer. Today we are going to dismantle a problem that looks like a monster but is actually a masterpiece of mathematical elegance.

We start with the function

f (x) = e^{- x} sin x

and an integral function defined as:

F (x) = \int_{0}^{x} f (t) d t

At first glance, you might be tempted to dive into integration by parts, but stop! Pause! The JEE Advanced is not just about calculation; it is about observation.

The Leibniz Revelation

The first step is the Leibniz Rule. When you see

F (x) = \int_{0}^{x} f (t) d t

, your brain should immediately recognize the Fundamental Theorem of Calculus:

F^{'} (x) = f (x)

This is the key that unlocks the entire door. Many students get lost trying to solve the integral of

e^{- t} sin t

first, but that is a detour.

By substituting

F^{'} (x) = f (x)

directly into our target integral

I = \int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x

, we transform the expression into:

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

The Algebraic Dance

Now, look at the magic. Substituting

f (x) = e^{- x} sin x

leads to:

I = \int_{0}^{1} 2 (e^{- x} sin x) e^{x} d x

The exponential terms

e^{- x}

and

e^{x}

multiply to

e^{0} = 1

. They vanish! We are left with:

I = 2 \int_{0}^{1} sin x d x

This is the beauty of the problem—the complexity collapses into a simple sine integral. Evaluating this, we get:

I = 2 [- cos x]_{0}^{1} = 2 (1 - cos 1)

The Taylor Series Bridge

But wait, how do we evaluate

cos 1

? We can approximate it using the Taylor series:

cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

By plugging in

x = 1

, we get

cos 1 \approx 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720}

When we substitute this back into our expression

I = 2 (1 - cos 1)

, the leading

1

cancels out, and we are left with a series that we can bound.

By calculating the first few terms, we find the value lies between

\frac{330}{360}

and

\frac{331}{360}

. This is the essence of JEE physics and math: simplify, observe, and conquer. You didn't need a calculator; you needed insight.

Common Traps & FAQs

Wait, why is

F^{'} (x) = f (x)

? I thought I had to integrate

f (t)

first?

That is a classic trap! The Fundamental Theorem of Calculus (Leibniz Rule) tells us that the derivative of the integral

\int_{0}^{x} f (t) d t

is simply the integrand

f (x)

. You do not need to perform the integration; you just need to recognize the relationship. It is a gift from the problem setter to save you time!

Why can't I just use integration by parts for the whole thing?

You absolutely could, but it would be like using a sledgehammer to crack a nut. The problem is designed to reward those who spot the symmetry and the relationship between

F^{'} (x)

and

f (x)

first. Always look for the 'elegant path' before starting the 'brute force path'.

How do I know how many terms of the Taylor series to take?

Great question! Since this is an alternating series, the error is always less than the absolute value of the first neglected term. By taking three terms, we get a very tight bound that fits perfectly into the options provided. In JEE, usually, three to four terms are more than enough to narrow down the range.

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be defined as $f (x) = e^{- x} sin x$ . If $F : [0, 1] \to R$ is a differentiable function such that $F (x) = \int_{0}^{x} f (t) d t$ , then the value of $\int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x$ lies in the interval

Select Answer:

Visualized Solution

Defining the Functions

Fundamental Theorem of Calculus

Differentiating $F (x)$

Substituting into the Target Integral

Simplifying the Integrand

Substituting $f (x)$

Canceling Exponential Terms

Visualizing the Definite Integral

Integrating Sine Function

Applying the Limits

Taylor Series Expansion

Expanding $cos 1$

Substituting Series into Integral

Simplifying the Series

Bounding the Alternating Series

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Leibniz Revelation

The Algebraic Dance

The Taylor Series Bridge

Similar Questions

Let $f$ be a differentiable function in $(0, \frac{π}{2})$ . If $\int_{c o s x}^{1} t^{2} f (t) d t = sin^{3} x + cos x$ , then $\frac{1}{3} f^{'} (\frac{1}{3})$ is equal to :

Let $F (x) = \int_{x}^{x^{2} + \frac{π}{6}} 2 cos^{2} t (d t)$ for all $x \in R$ and $f : [0, 1/2] \to [0, \infty)$ be a continuous function. For $α \in [0, 1/2]$ , if $F^{'} (α) + 2$ is the area of the region bounded by $x = 0, y = 0, y = f (x)$ and $x = α$ , then $f (0)$ is

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

If $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :

Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then

For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then

Select Answer:

Visualized Solution

Defining the Functions

Fundamental Theorem of Calculus

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(x)

Substituting into the Target Integral

Simplifying the Integrand

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

Canceling Exponential Terms

Visualizing the Definite Integral

Integrating Sine Function

Applying the Limits

Taylor Series Expansion

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos1

Substituting Series into Integral

Simplifying the Series

Bounding the Alternating Series

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Leibniz Revelation

The Algebraic Dance

The Taylor Series Bridge

Similar Questions

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Differentiating $F (x)$

Substituting $f (x)$

Expanding $cos 1$

Let $f$ be a differentiable function in $(0, \frac{π}{2})$ . If $\int_{c o s x}^{1} t^{2} f (t) d t = sin^{3} x + cos x$ , then $\frac{1}{3} f^{'} (\frac{1}{3})$ is equal to :

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

If $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :

Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then

For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then