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MCQMediumJEE 2023Applications of Integrals (Area)

JEE Mathematics 2023 Question with Solution

The area of the region given by {(x,y):xy≤8, 1≤y≤x2}\{(x, y) : xy \leq 8, \, 1 \leq y \leq x^2\} is:

  • A

    8ln⁡e2−1338 \ln_e 2 - \frac{13}{3}

  • B

    16ln⁡e2−14316 \ln_e 2 - \frac{14}{3}

  • C

    8ln⁡e2+768 \ln_e 2 + \frac{7}{6}

  • D

    16ln⁡e2+7316 \ln_e 2 + \frac{7}{3}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The region is defined by

xy≤8,1≤y≤x2xy \leq 8, \quad 1 \leq y \leq x^2

Find: The area of the region.

The boundaries of the region are y=1y = 1, y=x2y = x^2, and y=8xy = \frac{8}{x}.

The region is split into two parts:

  • For x∈[1,2]x \in [1,2], the area is between y=x2y = x^2 and y=1y = 1.
  • For x∈[2,8]x \in [2,8], the area is between y=8xy = \frac{8}{x} and y=1y = 1.

So,

Area=∫12(x2−1) dx+∫28(8x−1) dx\text{Area} = \int_1^2 (x^2 - 1) \, dx + \int_2^8 \left(\frac{8}{x} - 1\right) \, dx

Now evaluate the first integral:

∫12(x2−1) dx=∫12x2 dx−∫121 dx\int_1^2 (x^2 - 1) \, dx = \int_1^2 x^2 \, dx - \int_1^2 1 \, dx ∫12x2 dx=[x33]12=83−13=73\int_1^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} ∫121 dx=[x]12=2−1=1\int_1^2 1 \, dx = \left[x\right]_1^2 = 2 - 1 = 1 ∫12(x2−1) dx=73−1=43\int_1^2 (x^2 - 1) \, dx = \frac{7}{3} - 1 = \frac{4}{3}

Now evaluate the second integral:

∫28(8x−1) dx=∫288x dx−∫281 dx\int_2^8 \left(\frac{8}{x} - 1\right) \, dx = \int_2^8 \frac{8}{x} \, dx - \int_2^8 1 \, dx ∫288x dx=8∫281x dx=8[ln⁡x]28=8(ln⁡8−ln⁡2)=8ln⁡4=16ln⁡2\int_2^8 \frac{8}{x} \, dx = 8 \int_2^8 \frac{1}{x} \, dx = 8[\ln x]_2^8 = 8(\ln 8 - \ln 2) = 8\ln 4 = 16\ln 2 ∫281 dx=[x]28=8−2=6\int_2^8 1 \, dx = \left[x\right]_2^8 = 8 - 2 = 6 ∫28(8x−1) dx=16ln⁡2−6\int_2^8 \left(\frac{8}{x} - 1\right) \, dx = 16\ln 2 - 6

Add both parts:

Area=43+(16ln⁡2−6)\text{Area} = \frac{4}{3} + (16\ln 2 - 6) Area=16ln⁡2−6+43=16ln⁡2−183+43=16ln⁡2−143\text{Area} = 16\ln 2 - 6 + \frac{4}{3} = 16\ln 2 - \frac{18}{3} + \frac{4}{3} = 16\ln 2 - \frac{14}{3}

Therefore, the area of the region is 16ln⁡2−14316\ln 2 - \frac{14}{3}.

The solution states the correct option as A, but this disagrees with the computed value and the listed options. The computed value matches option B.

Common mistakes

  • Taking the upper boundary as y=x2y = x^2 for all values of xx is incorrect because the condition xy≤8xy \leq 8 also imposes y≤8xy \leq \frac{8}{x}. Compare the two upper curves and split the region where they intersect.

  • Using the wrong intersection point of y=x2y = x^2 and y=8xy = \frac{8}{x} leads to incorrect limits. Solve x2=8xx^2 = \frac{8}{x} carefully to get x3=8x^3 = 8, so the split occurs at x=2x = 2.

  • Forgetting the lower boundary y=1y = 1 gives an overestimate of area. Each vertical strip must be computed as upper function minus lower function, not just the upper curve alone.

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