The area of the region {(x, y) : xy ≤ 8, 1 ≤ y ≤ x²} is (1) 8log_e 2 - 14/3 (2) 16log_e 2 - 14/3 (3)

Q: Why do we have to split the integral at $x=2$?

Think of the area as a room. The 'roof' of this room is defined by the upper boundary. From $x=1$ to $x=2$, the parabola $y=x^2$ is the ceiling. But as you move past $x=2$, the hyperbola $y=\frac{8}{x}$ dips down and becomes the new ceiling. Since the function defining the upper boundary changes, you cannot describe the entire area with a single integral. You must split the journey at the point where the ceiling changes!

Q: Can I integrate with respect to $y$ instead of $x$?

That is a brilliant question! Yes, you absolutely can. You would express $x$ in terms of $y$ ($x=\sqrt{y}$ and $x=\frac{8}{y}$) and integrate along the y-axis. It is a fantastic way to verify your answer. However, in this specific problem, integrating with respect to $x$ is more direct because the lower boundary is a simple horizontal line $y=1$.

Q: Why did we ignore the negative intersection points?

Great observation. While the algebraic equations $x^2=1$ or $x^3=8$ might suggest other solutions, we are looking for a physical region in the first quadrant. The constraints $1 \le y \le x^2$ and $xy \le 8$ naturally restrict us to the positive coordinate plane. Always keep the physical context of the problem in mind!

Question

The area of the region $\{(x, y) : xy \le 8, 1 \le y \le x^2\}$ is

Options:

(A) $8\log_e 2 - \frac{14}{3}$

(B) $16\log_e 2 - \frac{14}{3}$

(C) $16\log_e 2 - 6$

(D) $8\log_e 2 - \frac{7}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of the Landscape

Welcome, fellow explorer of the mathematical universe! Today, we are not just solving an integral; we are mapping a landscape.

Imagine you are standing on a coordinate plane. You have three distinct paths defining a region: a flat floor at $y=1$, a rising parabolic wall $y=x^2$, and a descending hyperbolic ceiling $y=\frac{8}{x}$.

Our mission is to calculate the exact area of the space enclosed by these three curves. This is a classic JEE problem because it tests your ability to visualize geometry before you even touch a pen to paper.

### The Crossroads: Finding the Intersection Points

Before we can integrate, we must know where our journey begins and ends. We need the 'corner points' of our region.

First, where does the parabola $y=x^2$ meet the floor $y=1$? Setting $x^2=1$, we find $x=1$. So, our journey starts at $(1, 1)$.

Next, where does the parabola meet the hyperbola? We set $x^2 = \frac{8}{x}$, which leads us to $x^3=8$, giving us $x=2$. At $x=2$, $y=4$. This is the peak of our region at $(2, 4)$.

Finally, where does the hyperbola meet the floor? Setting $\frac{8}{x}=1$, we find $x=8$. Our journey ends at $(8, 1)$. We have our coordinates: $(1, 1)$, $(2, 4)$, and $(8, 1)$.

### The Divide: Why One Integral is Not Enough

Here is where many students stumble. They try to write one single integral for the whole area.

But look at the 'roof' of our region. From $x=1$ to $x=2$, the roof is the parabola $y=x^2$. But as soon as we cross $x=2$, the hyperbola $y=\frac{8}{x}$ takes over as the roof.

Because the upper boundary function changes, we must split our integral into two distinct parts. This is the heart of the problem: recognizing that the 'height' of our vertical strips changes its definition at $x=2$.

### The Calculation: Bringing it Home

Let us set up our two integrals. For the first part, from $x=1$ to $x=2$, the height of our strip is the upper curve minus the lower curve: $(x^2 - 1)$.

$$A_1 = \int_1^2 (x^2 - 1) dx$$

Integrating this, we get:

$$\left[ \frac{x^3}{3} - x \right]_1^2 = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 1 \right) = \frac{2}{3} - \left( -\frac{2}{3} \right) = \frac{4}{3}$$

Now, for the second part, from $x=2$ to $x=8$, the height is $(\frac{8}{x} - 1)$.

$$A_2 = \int_2^8 \left( \frac{8}{x} - 1 \right) dx$$

Integrating this, we get:

$$\left[ 8\ln x - x \right]_2^8 = (8\ln 8 - 8) - (8\ln 2 - 2)$$

Since $8\ln 8 = 8(3\ln 2) = 24\ln 2$, this simplifies to:

$$24\ln 2 - 8 - 8\ln 2 + 2 = 16\ln 2 - 6$$

### The Final Synthesis

We are almost there! We simply add our two areas together:

$$A = A_1 + A_2 = \frac{4}{3} + 16\ln 2 - 6$$

Combining the constants, we have:

$$\frac{4}{3} - 6 = \frac{4-18}{3} = -\frac{14}{3}$$

Thus, our final area is **$16\ln 2 - \frac{14}{3}$**.

Take a moment to appreciate this result. It is not just a number; it is the exact measure of the space we defined. You have successfully navigated the curves, split the integral, and arrived at the truth.

Animated Solution for Mathematics - Definite Integration: The area of the region ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is

Select Answer:

Visualized Solution

Visualizing the Boundary Curves

Identifying the Bounded Region

Intersection: Parabola and Line

Intersection: Parabola and Hyperbola

Intersection: Hyperbola and Line

Splitting the Region

Setting up Area 1

Setting up Area 2

Total Area Expression

Integrating the First Part

Integrating the Second Part

Applying Limits

Simplifying the Expression

Final Answer

The Sigma Insight: Area Bounded by Curves

The Geometry of the Landscape

The Crossroads

The Divide

The Calculation

The Final Synthesis

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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):xy≤8,1≤y≤x2} is