The area (in sq. units) of the region bounded by the parabola, y = x² + 2 and the lines, y = x + 1,

Q: Wait, why do we subtract the line from the parabola? Could it be the other way around?

Great question! The area between two curves must always be a positive value. By definition, the area is the integral of the 'upper' curve minus the 'lower' curve. If you subtract them in the wrong order, you will get a negative result. Always check which function is greater in the given interval—in this case, $x^2 + 2$ is clearly above $x + 1$ for $x \in [0, 3]$.

Q: Do I need to check for intersection points between $x=0$ and $x=3$?

That is a very sharp observation! Yes, in general, you must always check if the curves intersect within your interval of integration. If they cross, the 'upper' and 'lower' functions would swap, and you would need to split the integral into two parts. Here, $x^2 + 2 = x + 1$ leads to $x^2 - x + 1 = 0$, which has no real roots (discriminant $D = 1 - 4 = -3 < 0$). Since they never cross, our single integral is perfectly valid.

Q: Why is the lower limit zero? Does it always happen?

Not at all! The limits of integration are determined by the boundaries provided in the problem. Here, the problem explicitly states the region is bounded by the lines $x = 0$ and $x = 3$. If the problem had asked for the area bounded *only* by the parabola and the line, you would have had to find the intersection points to determine the limits yourself. Always read the problem statement carefully to identify your boundaries.

Question

The area (in sq. units) of the region bounded by the parabola, $y = x^2 + 2$ and the lines, $y = x + 1$, $x = 0$ and $x = 3$, is :

Options:

(A) $15/4$

(B) $15/2$

(C) $21/2$

(D) $17/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a vast, flat, mathematical landscape. Before you, two distinct paths are carved into the earth. One is a graceful, upward-opening parabola defined by the equation $y = x^2 + 2$.

Its vertex, the lowest point of its journey, rests elegantly at $(0, 2)$ on the y-axis, never dipping down to touch the x-axis. The other path is a straight, unwavering line, $y = x + 1$, cutting across the plane with a steady slope of $1$.

Our goal is to calculate the precise area of the region trapped between these two paths, enclosed by the vertical walls at $x = 0$ and $x = 3$.

### Visualizing the Landscape

Before we touch a single integral sign, we must see the region. When you plot $y = x^2 + 2$ and $y = x + 1$, you will notice that for every value of $x$ between $0$ and $3$, the parabola is flying higher than the line.

It acts as a protective ceiling, while the line serves as a solid floor. This is the 'upper' and 'lower' relationship that defines our area.

By seeing the curves, we understand that our area is simply the accumulation of the vertical distance between the ceiling and the floor at every point along the x-axis.

### The Calculus Engine

We use the power of the definite integral to sum up this area. The area $A$ is the integral of the difference between the upper function and the lower function:

$$A = \int_{0}^{3} [y_{upper} - y_{lower}] dx$$

Substituting our functions, we get:

$$A = \int_{0}^{3} [(x^2 + 2) - (x + 1)] dx$$

Before we integrate, we simplify the integrand. Subtracting the line from the parabola, we get $x^2 + 2 - x - 1$, which simplifies to $x^2 - x + 1$. Now, our problem is transformed into the integral of a simple polynomial:

$$A = \int_{0}^{3} (x^2 - x + 1) dx$$

### The Final Ascent

We apply the fundamental theorem of calculus by integrating term by term using the power rule. The antiderivative is:

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

Plugging in the upper limit $x = 3$, we calculate:

$$\left( \frac{3^3}{3} - \frac{3^2}{2} + 3 \right) = \left( \frac{27}{3} - \frac{9}{2} + 3 \right) = 9 - 4.5 + 3 = 7.5$$

Plugging in the lower limit $x = 0$ yields $0$. Subtracting the lower limit from the upper limit, we arrive at $7.5 - 0 = 7.5$.

Converting this to a fraction, we find our final answer: **$\frac{15}{2}$ square units**.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region bounded by the parabola, $y = x^{2} + 2$ and the lines, $y = x + 1$ , $x = 0$ and $x = 3$ , is :

Select Answer:

Visualized Solution

Visualize the Parabola $y = x^{2} + 2$

Plot the Line $y = x + 1$

Define the Vertical Bounds $x = 0$ and $x = 3$

Identify the Region of Interest

Set up the Area Integral

Substitute the Functions

Simplify the Integrand

Apply the Power Rule of Integration

Evaluate the Upper Limit $x = 3$

Evaluate the Lower Limit $x = 0$

Final Calculation and Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Visualizing the Landscape

The Calculus Engine

The Final Ascent

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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 bounded by the parabola, y=x2+2 and the lines, y=x+1, x=0 and x=3, is :

Select Answer:

Visualized Solution

Visualize the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x2+2

Plot the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x+1

Define the Vertical Bounds .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3