If the area of the region bounded by the curves y=4-x^2/4 and y=(x-4)/2 is equal to alpha, then 6alpha equals

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2025

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Home Maths IntegralsQuestion #223

If the area of the region bounded by the curves

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

and

y = \frac{x - 4}{2}

is equal to

α

, then

6 α

equals

Clue 1 of 3: Area Between Curves

The area between two curves $y = f (x)$ and $y = g (x)$ over an interval $[a, b]$ is given by the integral of the absolute difference of the functions.

🧠 Full Solution Path

Step 1: Find Intersection Points✦ Active

Set the equations of the curves equal to each other to find the x-coordinates of the intersection points.

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

Multiplying by 4, we get $16 - x^{2} = 2 (x - 4)$ , which simplifies to $x^{2} + 2 x - 24 = 0$ . Factoring the quadratic equation gives $(x + 6) (x - 4) = 0$ . Thus, the intersection points are $x = - 6$ and $x = 4$ . These will be the limits of integration.

Step 2: Determine Upper and Lower Curves and Set up Integral○ Expand

Let $y_{1} = 4 - \frac{x^{2}}{4}$ (parabola) and $y_{2} = \frac{x - 4}{2}$ (line). Test a point within the interval $[- 6, 4]$ , for example, $x = 0$ . $y_{1} (0) = 4$ and $y_{2} (0) = - 2$ . Since $y_{1} (0) > y_{2} (0)$ , the parabola is the upper curve. The area $α$ is given by the integral:

α = \int_{- 6}^{4} ((4 - \frac{x^{2}}{4}) - (\frac{x - 4}{2})) d x

Simplify the integrand:

α = \int_{- 6}^{4} (6 - \frac{x}{2} - \frac{x^{2}}{4}) d x

💡 Teacher's Secret Hint

Carefully expand and combine terms before integrating to avoid errors.

Step 3: Evaluate the Definite Integral and Calculate

6 α

○ Expand

Integrate the expression:

α = {[6 x - \frac{x^{2}}{4} - \frac{x^{3}}{12}]}_{- 6}^{4}

Substitute the limits of integration:

α = (6 (4) - \frac{4^{2}}{4} - \frac{4^{3}}{12}) - (6 (- 6) - \frac{(- 6)^{2}}{4} - \frac{(- 6)^{3}}{12})

α = (24 - 4 - \frac{64}{12}) - (- 36 - \frac{36}{4} - \frac{- 216}{12})

α = (20 - \frac{16}{3}) - (- 36 - 9 + 18)

α = (\frac{60 - 16}{3}) - (- 27)

α = \frac{44}{3} + 27 = \frac{44 + 81}{3} = \frac{125}{3}

Finally, calculate $6 α$ :

6 α = 6 \times \frac{125}{3} = 2 \times 125 = 250

💡 Teacher's Secret Hint

Double-check arithmetic, especially with negative signs and fractions during substitution.

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Maths Question 19 – JEE-MAIN 2025

Clue 1 of 3: Area Between Curves

🧠 Full Solution Path