Region Bounded by Curves Shortcut

 

❓ Question

Find the area bounded by:

$$\{(x,y): xy\le 8,\ y\le x^2,\ y\ge1\}$$

Options:

1. $\displaystyle 16\log_e2-\frac{14}{3}$
2. $\displaystyle 16\log_e2-\frac{13}{3}$
3. $\displaystyle 16\log_e2-\frac{17}{3}$
4. $\displaystyle 16\log_e2-\frac{19}{3}$

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Area bounded by curves
👉 Region inequalities
👉 Definite integration.

Main idea:

Convert inequalities into curves:

$$xy\le8 \Rightarrow y\le\frac8x$$
$$y\le x^2$$
$$y\ge1$$

Then identify the bounded region carefully.

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🔷 Step 1 — Understand the Region 💯

Given:

$$y\le\frac8x$$
$$y\le x^2$$
$$y\ge1$$

Since:

$$y\ge1$$

region lies above the line:

$$y=1$$

Upper boundary is the smaller of:

$$x^2 \quad\text{and}\quad \frac8x$$

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🔷 Step 2 — Find Intersection Points

Between $y=x^2$ and $y=1$

$$x^2=1$$
$$x=\pm1$$

Relevant positive side:

$$x=1$$

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Between $y=\frac8x$ and $y=1$

$$\frac8x=1$$
$$x=8$$

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Between $y=x^2$ and $y=\frac8x$

$$x^2=\frac8x$$
$$x^3=8$$
$$x=2$$

and:

$$y=4$$

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🔷 Step 3 — Split Region

For:

$$1\le x\le2$$

upper curve is:

$$y=x^2$$

For:

$$2\le x\le8$$

upper curve is:

$$y=\frac8x$$

Lower curve throughout:

$$y=1$$

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🔷 Step 4 — Form Area Integral

Thus:

$$A= \int_1^2(x^2-1)\,dx + \int_2^8\left(\frac8x-1\right)\,dx$$

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🔷 Step 5 — Evaluate First Integral

$$\int_1^2(x^2-1)\,dx = \left[\frac{x^3}{3}-x\right]_1^2$$
$$= \left(\frac83-2\right)-\left(\frac13-1\right)$$
$$= \frac23+\frac23$$
$$= \frac43$$

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🔷 Step 6 — Evaluate Second Integral

$$\int_2^8\left(\frac8x-1\right)\,dx = \left[8\ln x-x\right]_2^8$$
$$= (8\ln8-8)-(8\ln2-2)$$
$$= 8(\ln8-\ln2)-6$$
$$= 8\ln4-6$$
$$= 16\ln2-6$$

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🔷 Step 7 — Final Area

$$A = \frac43+16\ln2-6$$
$$= 16\ln2-\frac{14}{3}$$

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✅ Final Answer

$$\boxed{ 16\log_e2-\frac{14}{3} }$$

(Option 1)

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🔷 JEE Trap Alert 🚨

❌ Upper curve identify galat kar dena

Always compare:

$$x^2 \quad\text{and}\quad \frac8x$$

around intersection point $x=2$.

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