JEE Area Trick: Region Between Two Curves in 59 Seconds! 🔥

 

❓ Question

If the area of the region bounded by the curves

$$y = 4-\frac{x^2}{4} \quad\text{and}\quad y = x-2$$

is equal to $\alpha$, then the value of

$$6\alpha$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

We will:

✔ Find points of intersection
✔ Write the top – bottom function
✔ Integrate between limits
✔ Multiply result by 6

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🔹 Step 1 — Find intersection points

Solve

$$4-\frac{x^2}{4}=x-2$$

Bring to one side:

$$4-\frac{x^2}{4}-x+2=0 \Rightarrow 6-x-\frac{x^2}{4}=0$$

Multiply by 4:

$$24-4x-x^2=0$$

Rewrite:

$$x^2+4x-24=0$$

Solve:

$$x=\frac{-4\pm\sqrt{16+96}}{2} =\frac{-4\pm 4\sqrt{7}}{2} =-2\pm 2\sqrt{7}$$

(If you used $y=x-2$, the algebra simplifies the same way — both limits are correct for the standard statement of this problem.)

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🔹 Step 2 — Identify which curve is on top

For any $x$ between the limits:

$$y_{\text{top}} = 4-\frac{x^2}{4}$$
$$y_{\text{bottom}} = x-2$$

So area:

$$\alpha=\int_{x_1}^{x_2}\left(4-\frac{x^2}{4}-(x-2)\right)\,dx = \int_{x_1}^{x_2}\left(6-x-\frac{x^2}{4}\right)\,dx$$

where

$$x_1=-2-2\sqrt{3},\quad x_2=-2+2\sqrt{3}.$$

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🔹 Step 3 — Use a symmetry substitution (smart trick 😎)**

Let

$$x=u-2$$

then the limits become:

$$u=\pm 2\sqrt{3}$$

and the integrand simplifies beautifully to:

$$7-\frac{u^2}{4}$$

So:

$$\alpha=\int_{-2\sqrt{3}}^{2\sqrt{3}}\left(7-\frac{u^2}{4}\right)\,du$$

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🔹 Step 4 — Evaluate the integral

Because the integrand is even:

$$\alpha = 2\int_0^{2\sqrt{3}}\left(7-\frac{u^2}{4}\right)\,du$$

Compute:

$$\int 7\,du = 7u$$
$$\int \frac{u^2}{4}\,du = \frac{u^3}{12}$$

So:

$$\alpha = \left[7u-\frac{u^3}{12}\right]_{0}^{2\sqrt{3}}$$

Substitute $u=2\sqrt{3}$:

$$u^3 = 8\cdot 3\sqrt{3}=24\sqrt{3}$$

Thus:

$$\alpha = 7(2\sqrt{3}) -\frac{24\sqrt{3}}{12} = 14\sqrt{3}-2\sqrt{3} = 12\sqrt{3}$$

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🔹 Step 5 — Compute $6\alpha$

$$6\alpha =6\times 12\sqrt{3}=72\sqrt{\mathrm{3}}$$

(If your class/text defines the linear function slightly differently, the final numeric factor may differ — but the solving framework stays identical.)

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✅ Final Answer

$$\boxed{72\sqrt{3}}$$