Let the area enclosed by the lines x + y = 2, y = 0, x = 0 and the curve f(x) = min{x² + 3/4, 1 + [x

Q: Wait, why do we split the integral at $x=1/2$?

Great question! The function $f(x)$ is defined by a 'minimum' operator. At $x=1/2$, the parabola $x^2 + 3/4$ hits the value $1$. Before $1/2$, the parabola is lower; after $1/2$, the constant $1$ is lower. We must follow the 'floor' of the function to correctly define the area.

Q: Why does the upper boundary change to $2-x$ in the second interval?

Look at the geometry. The line $x+y=2$ (or $y=2-x$) is a boundary of the region. In the interval $[1, 2]$, the function $f(x)$ is $1 + [x] = 2$. But the line $y=2-x$ is *below* $2$ for $x > 1$. Since we want the area *enclosed* by these lines, the line $y=2-x$ acts as the 'roof' of our region.

Q: Is the greatest integer function always constant?

Yes, within any interval $[n, n+1)$, the value of $[x]$ is constant (it equals $n$). This is the secret to handling such functions—always break the domain into these integer-bounded intervals to turn the 'staircase' into a simple constant.

Question

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f(x) = \min\{x^2 + \frac{3}{4}, 1 + [x]\}$ where $[x]$ denotes the greatest integer $\le x$, be $A$. Then the value of $12A$ is

Sigma Percentile · Accepted Answer

### The Anatomy of the Function

Welcome, future engineers! Today, we are going to dissect a problem that looks intimidating but is actually a masterclass in geometric intuition. We are dealing with the area enclosed by $x+y=2$, $y=0$, $x=0$, and the curve $f(x) = \min\{x^2 + \frac{3}{4}, 1 + [x]\}$.

The key to solving this is to stop seeing it as one scary equation and start seeing it as a collection of simple, manageable shapes.

### Phase 1: Deconstructing the Landscape

Let's look at $f(x)$. The 'min' function is a filter that tells us to take the smaller of the two provided values at any point $x$.

For $x \in [0, 1)$, we know $[x] = 0$. So, $f(x) = \min\{x^2 + \frac{3}{4}, 1\}$.

To find where these two meet, we set $x^2 + \frac{3}{4} = 1$, which gives $x^2 = \frac{1}{4}$, or $x = \frac{1}{2}$. This means from $x=0$ to $x=1/2$, the parabola $x^2 + \frac{3}{4}$ is the 'floor', and from $x=1/2$ to $x=1$, the constant $1$ is the 'floor'.

### Phase 2: The Geometry of Constraints

This is where the JEE Advanced trap lies. The region is bounded by $x+y=2$, which can be rewritten as $y = 2-x$.

In the interval $[1, 2]$, this line is decreasing from $1$ to $0$. Our function $f(x)$ is at least $1.75$ in this interval.

Since $2-x$ is always less than or equal to $1$ in this interval, the line $y = 2-x$ is strictly below our curve $f(x)$. Therefore, the line $y = 2-x$ becomes the upper boundary of our region for $x \in [1, 2]$.

### Phase 3: The Calculus of Summation

Now, we simply sum the areas. We have three distinct regions to calculate:

**Region 1 ($0 \le x \le 1/2$):** The area under the parabola $x^2 + \frac{3}{4}$.
$$A_1 = \int_0^{1/2} (x^2 + \frac{3}{4}) dx = \left[\frac{x^3}{3} + \frac{3x}{4}\right]_0^{1/2} = \frac{1}{24} + \frac{3}{8} = \frac{10}{24} = \frac{5}{12}$$

**Region 2 ($1/2 \le x \le 1$):** The area under the constant line $y=1$.
$$A_2 = \int_{1/2}^1 1 dx = [x]_{1/2}^1 = 1 - \frac{1}{2} = \frac{1}{2} = \frac{6}{12}$$

**Region 3 ($1 \le x \le 2$):** The area under the line $y=2-x$.
$$A_3 = \int_1^2 (2 - x) dx = \left[2x - \frac{x^2}{2}\right]_1^2 = (4 - 2) - (2 - \frac{1}{2}) = 2 - \frac{3}{2} = \frac{1}{2} = \frac{6}{12}$$

### The Grand Finale

Adding these up, the total area $A$ is:
$$A = \frac{5}{12} + \frac{6}{12} + \frac{6}{12} = \frac{17}{12}$$

The question asks for $12A$. Multiplying $\frac{17}{12}$ by $12$ gives us the final result of **17**.

By breaking the problem into logical, geometric pieces, we turned a complex function into a simple sum. Keep this mindset, and no integral will ever defeat you!

Animated Solution for Mathematics - Definite Integration: Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f (x) = min {x^{2} + \frac{3}{4}, 1 + [x]}$ where $[x]$ denotes the greatest integer $\leq x$ , be $A$ . Then the value of $12 A$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Boundaries

Analyzing $f (x)$ for $0 \leq x < 1$

Plotting $f (x)$ for $0 \leq x < 1$

Analyzing $f (x)$ for $1 \leq x < 2$

Comparing $f (x)$ with $y = 2 - x$

Setting up the Integrals

Calculating Area $A_{1}$

Calculating Area $A_{2}$

Calculating Area $A_{3}$

Final Summation and Result

The Sigma Insight: Area Bounded by Curves

The Anatomy of the Function

Phase 1

Phase 2

Phase 3

The Grand Finale

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Boundaries

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

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

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1≤x<2

Comparing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2−x

Setting up the Integrals

Calculating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Calculating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Calculating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3​