Let A1 be the bounded area enclosed by the curves y = x² + 2, x + y = 8 and y-axis that lies in the

Q: Why is the upper limit of integration 2 for both areas?

For $A_1$, we found the intersection of $y=x^2+2$ and $y=8-x$ by solving $x^2+2 = 8-x$, which gives $x^2+x-6=0$, leading to $x=2$ and $x=-3$. Since we are in the first quadrant, we use $x=2$. For $A_2$, the problem explicitly defines the boundary $x=2$, so we integrate up to that point.

Q: How do I know which curve is the 'upper' one?

You can test a point within the interval $[0, 2]$, such as $x=1$. For $A_1$, at $x=1$, the line $y=8-1=7$ is greater than the parabola $y=1^2+2=3$. Thus, the line is the upper curve. For $A_2$, at $x=1$, the parabola $y=1^2+2=3$ is greater than the curve $y=\sqrt{1}=1$. Thus, the parabola is the upper curve.

Q: Is there a trick to avoid the integration?

While you could use geometric formulas for simple shapes, these curves are non-linear, so calculus is the most robust and required method. However, always sketch the curves first; it prevents sign errors and helps you visualize the 'upper' and 'lower' functions instantly.

Question

Let $A_1$ be the bounded area enclosed by the curves $y = x^2 + 2, x + y = 8$ and $y$-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y = x^2 + 2, y^2 = x, x = 2$, and $y$-axis that lies in the first quadrant. Then $A_1 - A_2$ is equal to

Options:

(A) $\frac{2}{3}(3\sqrt{2} + 1)$

(B) $\frac{2}{3}(2\sqrt{2} + 1)$

(C) $\frac{2}{3}(\sqrt{2} + 1)$

(D) $\frac{2}{3}(4\sqrt{2} + 1)$

Sigma Percentile · Accepted Answer

### The Geometry of Curves: A Journey into Area

Welcome, fellow explorer of mathematics! Today, we are embarking on a journey to find the area enclosed by curves. This is not just about crunching numbers; it is about visualizing the dance between functions on a coordinate plane.

Imagine you are standing on the Cartesian plane, looking at the first quadrant. We have two distinct regions, $A_1$ and $A_2$, and our goal is to find the difference between them. Let us begin by setting the stage.

### Phase 1: Visualizing $A_1$ and the Hunt for the Intersection

Our first region, $A_1$, is defined by the parabola $y = x^2 + 2$, the line $x + y = 8$, and the $y$-axis. The parabola $y = x^2 + 2$ is a classic, shifted upwards by two units. The line $x + y = 8$ can be rewritten as $y = 8 - x$.

To find the area, we need to know where these two paths cross. By setting $x^2 + 2 = 8 - x$, we arrive at the quadratic equation:

$$x^2 + x - 6 = 0$$

Factoring this, we get $(x + 3)(x - 2) = 0$. Since we are restricted to the first quadrant, we ignore $x = -3$ and embrace $x = 2$. This intersection point is our upper limit of integration.

### Phase 2: Calculating $A_1$

Now, we apply the fundamental principle of area between curves: the integral of the upper curve minus the lower curve. In our interval $[0, 2]$, the line $y = 8 - x$ sits above the parabola $y = x^2 + 2$.

Thus, the integral is:

$$A_1 = \int_0^2 ((8 - x) - (x^2 + 2)) \, dx$$

Simplifying the integrand, we get $\int_0^2 (6 - x - x^2) \, dx$. Integrating term by term, we find:

$$\left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_0^2$$

Substituting the limits, we get $(12 - 2 - \frac{8}{3}) = 10 - \frac{8}{3} = \frac{22}{3}$. This is the area of our first region.

### Phase 3: Defining and Calculating $A_2$

Next, we turn our attention to $A_2$. This region is bounded by the same parabola $y = x^2 + 2$, the curve $y^2 = x$ (which, in the first quadrant, is $y = \sqrt{x}$), the vertical line $x = 2$, and the $y$-axis.

Here, the parabola is the upper boundary, and the root curve is the lower boundary. The integral is:

$$A_2 = \int_0^2 ((x^2 + 2) - \sqrt{x}) \, dx$$

Integrating this, we get:

$$\left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_0^2$$

Plugging in $x = 2$, we obtain:

$$\left( \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) \right) = \frac{8}{3} + \frac{12}{3} - \frac{4\sqrt{2}}{3} = \frac{20 - 4\sqrt{2}}{3}$$

### Phase 4: The Final Synthesis

We have reached the climax of our journey. We need to find $A_1 - A_2$. Substituting our values, we have:

$$\frac{22}{3} - \frac{20 - 4\sqrt{2}}{3}$$

Combining these, we get:

$$\frac{22 - 20 + 4\sqrt{2}}{3} = \frac{2 + 4\sqrt{2}}{3}$$

Factoring out a 2, we arrive at **$\frac{2}{3}(1 + 2\sqrt{2})$**. This matches our second option perfectly. Remember, the beauty of these problems lies in the clarity of your diagram and the precision of your integration.

Select Answer:

Visualized Solution

Visualizing the First Curve

Defining Region $A_{1}$

Finding the Intersection Point

Setting up the Integral for $A_{1}$

Calculating Area $A_{1}$

Defining Region $A_{2}$

Setting up the Integral for $A_{2}$

Calculating Area $A_{2}$

Calculating $A_{1} - A_{2}$

Final Answer

The Sigma Insight: Area Bounded by Curves

The Geometry of Curves

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Visualizing the First Curve

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

Finding the Intersection Point

Setting up the Integral for .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; } A1​

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

Setting up the Integral for .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; } A2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​−A2​