JEE Main 2022 (27 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f (x) = min {[x - 1], [x - 2], \dots, [x - 10]}$ where $[t]$ denotes the greatest integer $\leq t$ . Then $\int_{0}^{10} f (x) d x + \int_{0}^{10} (f (x))^{2} d x + \int_{0}^{10} ∣ f (x) ∣ d x$ is equal to \_\_\_\_.

Q: Do I really need to change the limits of integration when I substitute $u = x-10$? Can't I just integrate with respect to $x$?

You *could* technically keep $x$, but it makes the integration much harder. By substituting $u = x-10$, you shift the interval from $[0, 10]$ to $[-10, 0]$. This makes the step function $[u]$ much easier to handle because the jumps occur at integer values of $u$, which are now clearly defined in your new range.

Q: How do I integrate the greatest integer function $[u]$ over an interval like $[-10, 0]$? Is there a formula?

There isn't a single 'formula' like there is for $x^n$, but there is a beautiful geometric interpretation! The integral $\int_{n}^{n+1} [u] du$ is simply the value of the function on that interval, which is $n$. So, $\int_{-10}^{0} [u] du$ is just the sum of the values $-10, -9, \dots, -1$. It's just a summation problem in disguise!

Enter Numerical Value:

Visualized Solution

Understanding $f (x)$

Given $f (x) = min {[x - 1], [x - 2], \dots, [x - 10]}$
$[t]$ denotes the greatest integer $\leq t$ .

Inequality of Arguments

For any real number $x$ , we know:
$x - 1 > x - 2 > \dots > x - 10$

Simplifying $f (x)$

The greatest integer function is non-decreasing.
Therefore, $[x - 1] \geq [x - 2] \geq \dots \geq [x - 10]$
The minimum value is the last term: $f (x) = [x - 10]$

Setting Up the Integral Sum

We need to evaluate three integrals:
$I = \int_{0}^{10} f (x) d x + \int_{0}^{10} (f (x))^{2} d x + \int_{0}^{10} ∣ f (x) ∣ d x$
Substitute $f (x) = [x - 10]$

Applying Variable Substitution

Let $u = x - 10$
Differentiating both sides: $d u = d x$

Transforming the Limits

Lower limit: when $x = 0 ⟹ u = - 10$
Upper limit: when $x = 10 ⟹ u = 0$
The new interval is $[- 10, 0]$

The Transformed Integrals

$I_{1} = \int_{- 10}^{0} [u] d u$
$I_{2} = \int_{- 10}^{0} [u]^{2} d u$
$I_{3} = \int_{- 10}^{0} ∣ [u] ∣ d u$

Evaluating $I_{1}$

In $[- 10, 0]$ , $[u]$ takes integer values from $- 10$ to $- 1$ .
Each interval has length $1$ .
$I_{1} = (- 10) + (- 9) + \dots + (- 1) = - 55$

Evaluating $I_{2}$

$I_{2} = \int_{- 10}^{0} [u]^{2} d u$
Sum of squares: $(- 10)^{2} + (- 9)^{2} + \dots + (- 1)^{2}$
$I_{2} = 1^{2} + 2^{2} + \dots + 1 0^{2} = \frac{10 \times 11 \times 21}{6} = 385$

Evaluating $I_{3}$

$I_{3} = \int_{- 10}^{0} ∣ [u] ∣ d u$
Sum of absolute values: $∣ - 10∣ + ∣ - 9∣ + \dots + ∣ - 1∣$
$I_{3} = 10 + 9 + \dots + 1 = 55$

Final Summation

Total Sum $= I_{1} + I_{2} + I_{3}$
Sum $= - 55 + 385 + 55$
Final Answer: 385

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

The Beauty of Simplification

Imagine you are standing before a mountain of notation. You see

f (x) = min {[x - 1], [x - 2], \dots, [x - 10]}

It looks like a beast, doesn't it? But in JEE Advanced, the most intimidating problems often hide the most elegant solutions. Let's peel back the layers.

The Non-Decreasing Secret

First, let's look at the greatest integer function,

[t]

. It is a non-decreasing function, meaning if

a < b

, then

[a] \leq [b]

Now, look at our arguments:

x - 10 < x - 9 < \dots < x - 1

. Because of the non-decreasing property, it is guaranteed that

[x - 10] \leq [x - 9] \leq \dots \leq [x - 1]

The minimum of this set is clearly the last term:

[x - 10]

. Just like that, the 'min' operator vanishes!

The Power of Substitution

Now we face the integral:

I = \int_{0}^{10} [x - 10] d x + \int_{0}^{10} ([x - 10])^{2} d x + \int_{0}^{10} ∣ [x - 10] ∣ d x

Integrating

[x - 10]

directly is messy. Let's use the substitution

u = x - 10

When

x = 0

u = - 10

. When

x = 10

u = 0

. Our integral becomes:

I = \int_{- 10}^{0} [u] d u + \int_{- 10}^{0} [u]^{2} d u + \int_{- 10}^{0} ∣ [u] ∣ d u

Evaluating the Sums

This is where the magic happens. We are integrating a step function over the interval

[- 10, 0]

For

I_{1} = \int_{- 10}^{0} [u] d u

, we are summing the values of

[u]

on each interval

[k, k + 1)

. This gives us:

(- 10) + (- 9) + \dots + (- 1) = - 55

For

I_{2} = \int_{- 10}^{0} [u]^{2} d u

, we square those values:

(- 10)^{2} + (- 9)^{2} + \dots + (- 1)^{2}

. This is the sum of the first ten squares:

\frac{10 ( 11 ) ( 21 )}{6} = 385

For

I_{3} = \int_{- 10}^{0} ∣ [u] ∣ d u

, we take the absolute values:

10 + 9 + \dots + 1 = 55

The Grand Finale

Adding them up:

- 55 + 385 + 55

The

- 55

and

+ 55

cancel out, leaving us with 385.

It is elegant, it is precise, and it is beautiful. You have conquered the beast!

Common Traps & FAQs

Wait, why is the minimum of

[x - 1], [x - 2], \dots, [x - 10]

just

[x - 10]

? Shouldn't it be the other way around?

That is a fantastic question! Remember that the greatest integer function,

[t]

, is non-decreasing. This means if

a \leq b

, then

[a] \leq [b]

. Since

x - 10 < x - 9 < \dots < x - 1

, it follows that

[x - 10] \leq [x - 9] \leq \dots \leq [x - 1]

. Therefore, the smallest value in that set is indeed the one with the smallest argument, which is

[x - 10]

Do I really need to change the limits of integration when I substitute

u = x - 10

? Can't I just integrate with respect to

x

You *could* technically keep

x

, but it makes the integration much harder. By substituting

u = x - 10

, you shift the interval from

[0, 10]

[- 10, 0]

. This makes the step function

[u]

much easier to handle because the jumps occur at integer values of

u

, which are now clearly defined in your new range.

How do I integrate the greatest integer function

[u]

over an interval like

[- 10, 0]

? Is there a formula?

There isn't a single 'formula' like there is for

x^{n}

, but there is a beautiful geometric interpretation! The integral

\int_{n}^{n + 1} [u] d u

is simply the value of the function on that interval, which is

n

. So,

\int_{- 10}^{0} [u] d u

is just the sum of the values

- 10, - 9, \dots, - 1

. It's just a summation problem in disguise!

Animated Solution for Mathematics - Definite Integration: Let $f (x) = min {[x - 1], [x - 2], \dots, [x - 10]}$ where $[t]$ denotes the greatest integer $\leq t$ . Then $\int_{0}^{10} f (x) d x + \int_{0}^{10} (f (x))^{2} d x + \int_{0}^{10} ∣ f (x) ∣ d x$ is equal to \_\_\_\_.

Enter Numerical Value:

Visualized Solution

Understanding $f (x)$

Inequality of Arguments

Simplifying $f (x)$

Setting Up the Integral Sum

Applying Variable Substitution

Transforming the Limits

The Transformed Integrals

Evaluating $I_{1}$

Evaluating $I_{2}$

Evaluating $I_{3}$

Final Summation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of Simplification

The Non-Decreasing Secret

The Power of Substitution

Evaluating the Sums

The Grand Finale

Similar Questions

The value of $\int_{1}^{a} [x] f^{'} (x) d x, a > 1$ where $[x]$ denotes the greatest integer not exceeding $x$ is

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Enter Numerical Value:

Visualized Solution

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

Inequality of Arguments

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

Setting Up the Integral Sum

Applying Variable Substitution

Transforming the Limits

The Transformed Integrals

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I3​

Final Summation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of Simplification

The Non-Decreasing Secret

The Power of Substitution

Evaluating the Sums

The Grand Finale

Similar Questions

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Understanding $f (x)$

Simplifying $f (x)$

Evaluating $I_{1}$

Evaluating $I_{2}$

Evaluating $I_{3}$

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Let $[t]$ denote the greatest integer $\leq t$ . Then the value of $8 \cdot \int_{- 1/2}^{1} ([2 x] + ∣ x ∣) d x$ is .

If $[t]$ denotes the greatest integer $\leq t$ , then the value of $\int_{0}^{1} [2 x - ∣3 x^{2} - 5 x + 2∣ + 1] d x$ is:

Let $[t]$ denote the greatest integer function. If $\int_{0}^{2.4} [x^{2}] d x = α + β 2 + γ 3 + δ 5$ , then $α + β + γ + δ$ is equal to

If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

The value $9 \int_{0}^{9} [\frac{10 x}{x + 1}] d x$ , where $[t]$ denotes the greatest integer less than or equal to $t$ , is _____