QJEEM25Ma684d: JEE Main 2025 Mathematics - Class 12 JEEM25Ma684d Solution: The area of the region enclosed by the curves and is :

Consider the curves

$y = x^2 - 4x + 4 = (x-2)^2$

and

$y^2 = 16 - 8x.$

Notice that the second equation can be rewritten in terms of $x$ :

$8x = 16 - y^2 \quad \Longrightarrow \quad x = 2 - \frac{y^2}{8}.$

Step 1. Find the Intersection Points

To find the points where the curves intersect, substitute

$y = (x-2)^2$

into

$y^2 = 16 - 8x.$

Let

$u = x - 2 \quad \text{so that} \quad y = u^2.$

Then

$y^2 = u^4$

and

$x = u + 2.$

Substitute into the second curve:

$u^4 = 16 - 8(u+2).$

Simplify the right side:

$16 - 8(u+2) = 16 - 8u - 16 = -8u.$

Thus, the equation becomes

$u^4 + 8u = 0 \quad \Longrightarrow \quad u(u^4/ u? \,\, **{correcting factor}**).$

In fact, factor by taking out a common factor $u$ :

$u(u^3 + 8) = 0.$

Thus, either

$u = 0 \quad \text{or} \quad u^3 = -8.$

For $u = 0$ :

$x = u + 2 = 2, \quad y = u^2 = 0.$

For $u^3 = -8$ :

$u = -2, \quad \text{so} \quad x = -2 + 2 = 0, \quad y = (-2)^2 = 4.$

The curves intersect at the points $(2, 0)$ and $(0, 4).$

Step 2. Express the Curves in Terms of $y$

It is easier to integrate horizontally by expressing $x$ as a function of $y$ .

From the second curve:

$x = 2 - \frac{y^2}{8}.$

From the first curve, solving

$y = (x-2)^2$

for $x$ gives

$x-2 = \pm\sqrt{y}.$

Since at the intersection $(0, 4)$ the $x$ –value is less than 2, we take the negative branch:

$x = 2 - \sqrt{y}.$

Thus, for a fixed $y$ between 0 and 4, the left boundary is

$x_{\text{left}} = 2 - \sqrt{y},$

and the right boundary is

$x_{\text{right}} = 2 - \frac{y^2}{8}.$

Step 3. Set Up the Integral for the Area

The horizontal distance between the curves at a given $y$ is

$\Delta x = x_{\text{right}} - x_{\text{left}} = \left(2 - \frac{y^2}{8}\right) - \left(2 - \sqrt{y}\right) = \sqrt{y} - \frac{y^2}{8}.$

Integrate with respect to $y$ from $y = 0$ to $y = 4$ :

$A = \int_{0}^{4} \left(\sqrt{y} - \frac{y^2}{8}\right) \, dy.$

Step 4. Evaluate the Integral

Write the integral as

$A = \int_{0}^{4} y^{1/2}\, dy - \frac{1}{8} \int_{0}^{4} y^2\, dy.$

Compute each term:

For the first integral:

$\int y^{1/2}\, dy = \frac{2}{3} y^{3/2}.$

For the second integral:

$\int y^2\, dy = \frac{y^3}{3}.$

Thus,

$A = \left[\frac{2}{3} y^{3/2}\right]_0^4 - \frac{1}{8}\left[\frac{y^3}{3}\right]_0^4.$

Substitute $y = 4$ (note that at $y = 0$ both terms vanish):

Compute $4^{3/2}$ :

$4^{3/2} = \left( \sqrt{4} \right)^3 = 2^3 = 8.$

Compute the first term:

$\frac{2}{3} \times 8 = \frac{16}{3}.$

Compute the second term:

$\frac{1}{8} \cdot \frac{64}{3} = \frac{64}{24} = \frac{8}{3}.$

Therefore, the area is

$A = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}.$

Final Answer

The area of the region enclosed by the curves is

$\frac{8}{3}.$