The area of the region enclosed by the curves y = x^2 - 4x + 4 an… — Solution | JEE Mathematics

MCQMediumJEE 2025Applications of Integrals (Area)

The area of the region enclosed by the curves $y = x^2 - 4x + 4$ and $y^2 = 16 - 8x$ is:

A
$\frac{4}{3}$
B
$8$
C
$\frac{8}{3}$
D
$5$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The curves are $y = x^2 - 4x + 4$ and $y^2 = 16 - 8x$ .

Find: The area enclosed by these two curves.

First find the points of intersection. From $y^2 = 16 - 8x$ , substitute $y = x^2 - 4x + 4$ :

(x^2 - 4x + 4)^2 = 16 - 8x

Expanding:

x^4 - 8x^3 + 24x^2 - 32x + 16 = 16 - 8x

So,

x^4 - 8x^3 + 24x^2 - 24x = 0

Factorizing:

x(x^3 - 8x^2 + 24x - 24) = 0

Further,

x(x - 2)(x^2 - 6x + 12) = 0

Hence the real intersection points occur at $x = 0$ and $x = 2$ .

On the interval $0 \le x \le 2$ , the upper curve is $y = \sqrt{16 - 8x}$ and the lower curve is $y = x^2 - 4x + 4$ . Therefore, the required area is

A = \int_0^2 \left[\sqrt{16 - 8x} - (x^2 - 4x + 4)\right] \, dx

Detailed Evaluation

Evaluate the two integrals separately:

A = \int_0^2 \sqrt{16 - 8x} \, dx - \int_0^2 (x^2 - 4x + 4) \, dx

For the first integral, let

u = 16 - 8x

Then

du = -8 \, dx

and so

dx = -\frac{1}{8} \, du

Thus,

\int \sqrt{16 - 8x} \, dx = \int \sqrt{u}\left(-\frac{1}{8}\right) \, du = -\frac{1}{8} \cdot \frac{2}{3}u^{3/2} = -\frac{1}{12}(16 - 8x)^{3/2}

Evaluating from $0$ to $2$ gives

\int_0^2 \sqrt{16 - 8x} \, dx = \frac{64}{12} = \frac{16}{3}

Shifted Parabola Shortcut

Using the shifted form shown in the second approach, write $y = (x-2)^2$ and rewrite the second curve relative to the shifted axis. Then the enclosed region matches the standard parabola area form, giving

\text{Area} = \frac{16ab}{3}

With $a = \frac{1}{4}$ and $b = 2$ ,

\text{Area} = \frac{16 \times \frac{1}{4} \times 2}{3} = \frac{8}{3}

Therefore, the correct option is C.

Continuing the standard evaluation for completeness, the second integral is

\int_0^2 (x^2 - 4x + 4) \, dx = \left[\frac{x^3}{3} - 2x^2 + 4x\right]_0^2 = \frac{8}{3}

Hence,

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

Therefore, the area of the enclosed region is $\frac{8}{3}$ , so the correct option is C.

Common mistakes

Taking $y = -\sqrt{16 - 8x}$ as the upper curve on $0 \le x \le 2$ is wrong because the enclosed region here lies above the parabola $y = x^2 - 4x + 4$ and below the positive branch. Always identify the actual top and bottom curves before integrating.
Using incorrect limits of integration without first finding the intersection points is wrong because the area must be computed only over the interval where the two curves bound a closed region. First solve the intersection equations to get $x = 0$ and $x = 2$ .
Squaring or expanding incorrectly in $(x^2 - 4x + 4)^2$ leads to wrong intersection points and hence a wrong area. Expand carefully or keep the expression in factored form as long as possible before simplifying.

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