The area of the region {(x,y):x²+4x+2≤y≤|x+2|} is equal to
(1) 7
(2) 5
(3) 24/5
(4) 20/3

Q: Wait, why do we reject the negative root when solving $|X|^2 - |X| - 2 = 0$?

That is a fantastic question! When we factorized the equation, we got $(|X| - 2)(|X| + 1) = 0$. This gives us $|X| = 2$ or $|X| = -1$. By definition, the absolute value of any real number $|X|$ must be non-negative (i.e., $|X| \ge 0$). Therefore, $|X| = -1$ is physically impossible in the real number system. We discard it because it yields no real solutions.

Q: Why can't we just integrate the original functions directly without the substitution?

You absolutely can! The math will still work. However, as an educator, I want you to develop 'mathematical intuition.' The substitution $X = x+2$ isn't just a shortcut; it's a way to simplify the problem's structure. It turns a shifted parabola into a standard one centered at the origin, which reduces the chance of making arithmetic errors with the limits of integration. Always look for ways to make the problem 'cleaner' before you start calculating.

Q: How do I know for sure that the modulus function is the upper bound?

Great observation! To check this, pick a test point within the interval of intersection, say $X = 1$. For the modulus function, $y = |1| = 1$. For the parabola, $y = 1^2 - 2 = -1$. Since $1 > -1$, the modulus function is clearly above the parabola in this region. Always perform a quick 'sanity check' like this before setting up your integral!

Question

The area of the region $\{(x,y):x^{2}+4x+2\le y\le|x+2|\}$ is equal to

Options:

(A) $7$

(B) $5$

(C) $24/5$

(D) $20/3$

Sigma Percentile · Accepted Answer

### The Art of Seeing Symmetry

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an area problem; we are learning to see the hidden elegance within algebraic expressions.

We are tasked with finding the area bounded by the parabola $y = x^2+4x+2$ and the absolute value function $y = |x+2|$. At first glance, this might look like a standard calculus problem, but the true JEE aspirant knows that the secret to speed and accuracy lies in simplification.

### The Power of the Shift

Look closely at the expressions. Do you see the term $x+2$ lurking in both?

The parabola $y = x^2+4x+2$ can be rewritten by completing the square: $y = (x^2+4x+4) - 2$, which is $y = (x+2)^2 - 2$.

Suddenly, the problem transforms. By substituting $X = x+2$, we shift our coordinate system. Our parabola becomes $y = X^2 - 2$, and our modulus function becomes $y = |X|$. We have moved the center of our universe to the origin, making the geometry symmetric and beautiful.

### Finding the Intersection

To find the boundaries of our region, we set the curves equal: $X^2 - 2 = |X|$.

Here is a pro-tip: remember that $X^2$ is identical to $|X|^2$. This allows us to write the equation as $|X|^2 - |X| - 2 = 0$.

This is a quadratic in disguise! Factoring this gives $(|X| - 2)(|X| + 1) = 0$. Since $|X|$ cannot be negative, we discard the $-1$ and find that $|X| = 2$, meaning our intersection points are at $X = 2$ and $X = -2$.

### The Elegance of Integration

Now, we visualize the region. Because both $y = X^2 - 2$ and $y = |X|$ are even functions (symmetric about the Y-axis), we don't need to integrate from $-2$ to $2$.

We can simply calculate the area from $0$ to $2$ and double it. This is the kind of strategic thinking that saves precious minutes in the exam hall.

In the interval $[0, 2]$, the modulus function $y = X$ is clearly above the parabola $y = X^2 - 2$. Thus, our integral becomes:

$$Area = 2 \int_{0}^{2} (X - (X^2 - 2)) \, dX$$

Simplifying the integrand, we get $X - X^2 + 2$. Integrating term by term, we have:

$$Area = 2 \left[ \frac{X^2}{2} - \frac{X^3}{3} + 2X \right]_{0}^{2}$$

Substituting the limits, we get:

$$Area = 2 \left( \left( \frac{4}{2} - \frac{8}{3} + 4 \right) - 0 \right) = 2 \left( 6 - \frac{8}{3} \right) = 2 \left( \frac{18-8}{3} \right) = \frac{20}{3}$$

And there it is! The **area is $\frac{20}{3}$ square units**. Remember, math is not just about the final number; it is about the journey of simplification and the beauty of symmetry.

Animated Solution for Mathematics - Definite Integration: The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq ∣ x + 2∣}$ is equal to

Select Answer:

Visualized Solution

Identify the Bounding Curves

Simplify using Substitution $X = x + 2$

Transforming the Parabola

Transforming the Modulus Function

Find Intersection Points

Solve the Quadratic Equation

Observe Symmetry

Set up the Definite Integral

Integrate the Expression

Evaluate the Limits

Final Calculation

The Sigma Insight: Area Bounded by Curves

The Art of Seeing Symmetry

The Power of the Shift

Finding the Intersection

The Elegance of Integration

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region {(x,y):x2+4x+2≤y≤∣x+2∣} is equal to