The area (in sq. units) of the region A = {(x, y) : x² ≤ y ≤ x + 2} is (1) 10/3 (2) 9/2 (3) 31/6 (4)

Q: Wait, how do I know which function is the 'upper' one without graphing it perfectly?

Great question! You can pick any test point between your intersection values. For example, at $x = 0$, the line gives $y = 2$ and the parabola gives $y = 0$. Since $2 > 0$, the line is clearly the upper boundary in that interval.

Q: Do I really need to use calculus? Can't I just use geometry formulas?

Geometry works for simple shapes like triangles or rectangles, but here, one boundary is a curve. Calculus is the tool that allows us to sum up an infinite number of tiny vertical strips to find the area under a curve—it is the only precise way to handle the curvature of the parabola.

Q: What happens if I accidentally subtract the functions in the wrong order?

If you subtract them in the wrong order, you will get a negative result. Remember, area represents a physical magnitude and must be positive. If you get a negative value, it simply means you swapped the upper and lower functions!

Q: Why are the intersection points $x = -1$ and $x = 2$ so important?

These points are the 'gatekeepers' of the region. They define exactly where the two curves meet and enclose the space. Outside these points, the curves diverge, and the region is no longer 'bounded' or 'trapped'.

Q: Is there a shortcut for the area between a parabola and a line?

Yes! For a parabola $y = ax^2 + bx + c$ and a line $y = mx + k$, the area is given by $\frac{|a|}{6}(\beta - \alpha)^3$, where $\alpha$ and $\beta$ are the roots. It is a beautiful shortcut, but always derive it via integration first to truly understand the physics of the problem!

Question

The area (in sq. units) of the region $A = \{(x, y) : x^2 \leq y \leq x + 2\}$ is

Options:

(A) $10/3$

(B) $9/2$

(C) $31/6$

(D) $13/6$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. Before you lies a landscape defined by two distinct mathematical entities: a graceful, upward-opening parabola, $y = x^2$, and a steady, rising line, $y = x + 2$.

These two paths are not merely lines on a page; they are boundaries that carve out a specific, finite piece of the plane. Our mission today is to measure the 'territory' trapped between them using the art of integral calculus.

### Finding the Gatekeepers

Before we can measure the area, we must identify the intersection points where the region begins and ends. We set the two functions equal to each other:

$$x^2 = x + 2$$

By rearranging this into a standard quadratic form, we obtain:

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

Factoring this equation gives us $(x - 2)(x + 1) = 0$. Our gatekeepers are revealed: the curves meet at $x = -1$ and $x = 2$. These two values serve as the limits of our integration.

### The Architecture of the Integral

To measure the region, we imagine filling the space between the curves with infinitely thin vertical rectangles. The height of each rectangle is the difference between the upper boundary (the line) and the lower boundary (the parabola).

The height of any slice at a position $x$ is defined as $(y_{upper} - y_{lower}) = (x + 2) - x^2$. To find the total area, we sum these infinite slices using the definite integral:

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

### The Power of Integration

Integration is the process of finding the anti-derivative. We integrate the expression term by term:

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

This is the moment of truth where we apply the Fundamental Theorem of Calculus to evaluate the definite integral.

### Final Calculation

We evaluate the expression at the upper limit ($x=2$) and subtract the value at the lower limit ($x=-1$).

At the upper limit, we calculate:

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

At the lower limit, we find:

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

Subtracting these values gives us the final area:

$$\frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6}$$

Simplifying this fraction, we arrive at our destination: **$\frac{9}{2}$ square units**.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region $A = {(x, y) : x^{2} \leq y \leq x + 2}$ is

Select Answer:

Visualized Solution

Visualize the Region $A$

Identify Intersection Points

Solve the Quadratic Equation

Define the Bounded Area

Set up the Definite Integral

Integrate the Expression

Substitute the Upper Limit

Substitute the Lower Limit

Calculate Final Area

Final Result

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Finding the Gatekeepers

The Architecture of the Integral

The Power of Integration

Final Calculation

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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 (in sq. units) of the region A={(x,y):x2≤y≤x+2} is