The area of the region bounded by y - x = 2 and x² = y is equal to :- (1) 16/3 (2) 2/3 (3) 9/2 (4) 4

Q: Why do we subtract the lower curve from the upper curve? Can't we just integrate both and add them?

That is a fantastic conceptual question! If you integrate both functions separately from -1 to 2, you are calculating the area under each curve all the way down to the x-axis. By subtracting the lower curve from the upper curve, you are effectively 'cutting out' the empty space under the parabola, leaving only the region trapped between the two lines. It is the geometric equivalent of subtraction.

Q: What happens if I accidentally swap the limits of integration?

If you swap the limits, you will get the exact same magnitude but with a negative sign. Remember, area is a physical quantity—it must be positive! If you ever get a negative result for an area problem, do not panic. It is a sign that you either swapped your limits or, more likely, you identified the 'upper' and 'lower' curves incorrectly. Simply take the absolute value or re-check your graph.

Q: Can I solve this by integrating with respect to $y$ instead of $x$?

You absolutely can! You would express $x$ in terms of $y$ for both curves ($x = \sqrt{y}$ and $x = y-2$). However, you would have to be very careful because the left boundary of the parabola changes at the vertex. Integrating with respect to $x$ is much cleaner here. In JEE, always look for the path of least resistance—choosing the right variable of integration is half the battle!

Question

The area of the region bounded by $y - x = 2$ and $x^2 = y$ is equal to :-

Options:

(A) $\frac{16}{3}$

(B) $\frac{2}{3}$

(C) $\frac{9}{2}$

(D) $\frac{4}{3}$

Sigma Percentile · Accepted Answer

### The Dance of Curves: Mastering Area Under a Curve

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey of visualization. In the world of JEE Advanced, the difference between a good student and a great one is the ability to see the geometry before writing a single line of algebra.

We are looking at the area bounded by a parabola, $y = x^2$, and a line, $y - x = 2$. This is a classic, elegant problem that tests your ability to bridge the gap between algebraic manipulation and geometric intuition.

### Phase 1: The Visualization

Imagine you are standing on a coordinate plane. You have a beautiful, symmetric parabola, $y = x^2$, opening its arms towards the sky. Now, imagine a straight line, $y = x + 2$, cutting through this parabola like a blade.

The region we are interested in is the 'pocket' trapped between the line and the curve. Before we do any calculus, we must know where this pocket begins and where it ends. We need the intersection points, where the $y$-values must be equal:

$$ x^2 = x + 2 $$

### Phase 2: The Algebraic Hunt

Now, let us solve this quadratic equation. We bring all terms to one side to set the equation to zero:

$$ x^2 - x - 2 = 0 $$

This is a standard quadratic. We are looking for two numbers that multiply to $-2$ and add to $-1$. Those numbers are $-2$ and $1$, allowing us to factorize:

$$ (x - 2)(x + 1) = 0 $$

This gives us our limits of integration: $x = -1$ and $x = 2$. These are the boundaries of our region. If you were to sketch this, you would see that the line enters the parabola at $x = -1$ and exits at $x = 2$.

### Phase 3: The Integral Setup

Here is where many students stumble. The rule is simple but profound: the area between two curves is the integral of the (Upper Curve) minus the (Lower Curve).

Looking at our sketch, between $x = -1$ and $x = 2$, the line $y = x + 2$ sits above the parabola $y = x^2$. Therefore, our integral becomes:

$$ A = \int_{-1}^{2} (x + 2 - x^2) dx $$

Do not let the integral sign intimidate you. It is just a fancy way of summing up an infinite number of tiny vertical strips, each with a height of $(x + 2 - x^2)$ and a width of $dx$.

### Phase 4: The Calculation

Now, we apply the Fundamental Theorem of Calculus. We integrate term by term:

1. The integral of $x$ is $\frac{x^2}{2}$.
2. The integral of $2$ is $2x$.
3. The integral of $x^2$ is $\frac{x^3}{3}$.

So, our antiderivative is:

$$ \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} ight]_{-1}^{2} $$

Now, we evaluate this at the upper limit ($x = 2$) and the lower limit ($x = -1$). For the upper limit ($x = 2$):

$$ \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} ight) = \left( 2 + 4 - \frac{8}{3} ight) = 6 - \frac{8}{3} = \frac{10}{3} $$

For the lower limit ($x = -1$):

$$ \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} ight) = \left( \frac{1}{2} - 2 + \frac{1}{3} ight) = \frac{3 - 12 + 2}{6} = -\frac{7}{6} $$

### The Final Triumph

Finally, we subtract the lower value from the upper value:

$$ A = \frac{10}{3} - \left( -\frac{7}{6} ight) = \frac{10}{3} + \frac{7}{6} $$

To add these, we find a common denominator of $6$:

$$ A = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} $$

Simplifying this fraction by dividing both numerator and denominator by $3$, we arrive at our beautiful final answer:

$$ A = \mathbf{\frac{9}{2} 	ext{ square units}} $$

There you have it! We started with a simple sketch, moved through the algebra of intersection, set up our integral with confidence, and arrived at the solution. Remember, calculus is not about memorizing formulas; it is about understanding the geometry of the world around you.

Animated Solution for Mathematics - Definite Integration: The area of the region bounded by $y - x = 2$ and $x^{2} = y$ is equal to :-

Select Answer:

Visualized Solution

Visualize the Curves

Finding Intersection Points

Solving the Quadratic Equation

Setting up the Integral

Integrating the Expression

Applying the Upper Limit

Applying the Lower Limit

Final Calculation

Key Takeaway

The Sigma Insight: Area Bounded by Curves

The Dance of Curves

Phase 1

Phase 2

Phase 3

Phase 4

The Final Triumph

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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 bounded by y−x=2 and x2=y is equal to :-