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Area Under the Curves question

2020 · 5 Sep · Shift 2 · Q27
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  5. /2020 · 5 Sep · Shift 2 · Q27

Area Under the Curves question

2020 · 5 Sep · Shift 2 · Q27

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region A = {(x, y) : (x – 1)[x] ≤\le≤ y ≤\le≤ 2 x\sqrt xx​, 0 ≤\le≤ x ≤\le≤ 2}, where [t] denotes the greatest integer function, is :
  1. A
    832−1{8 \over 3}\sqrt 2 - 138​2​−1
  2. B
    432+1{4 \over 3}\sqrt 2 + 134​2​+1
  3. C
    832−12{8 \over 3}\sqrt 2 - {1 \over 2}38​2​−21​
  4. D
    432−12{4 \over 3}\sqrt 2 - {1 \over 2}34​2​−21​
View written solutionFree

Correct answer: C

  1. We need the area of the region A={(x,y):(x−1)[x]≤y≤2x, 0≤x≤2}.A=\{(x,y):(x-1)[x]\le y\le 2\sqrt{x},\ 0\le x\le 2\}.A={(x,y):(x−1)[x]≤y≤2x​, 0≤x≤2}. So the required area is ∫02(2x−(x−1)[x]) dx.\int_0^2\Big(2\sqrt{x}-(x-1)[x]\Big)\,dx.∫02​(2x​−(x−1)[x])dx.

  2. Since [x][x][x] is the greatest integer function, on the interval [0,2][0,2][0,2] we have:

  • For 0≤x<10\le x<10≤x<1, [x]=0[x]=0[x]=0
  • For 1≤x<21\le x<21≤x<2, [x]=1[x]=1[x]=1
  • At x=2x=2x=2, [2]=2[2]=2[2]=2, but a single point does not affect area.

Hence,

0, & 0\le x<1,\\ x-1, & 1\le x<2. \end{cases}$$ 3. Therefore the area becomes $$\text{Area}=\int_0^1 2\sqrt{x}\,dx+\int_1^2\big(2\sqrt{x}-(x-1)\big)\,dx.$$ 4. Compute the first integral: $$\int_0^1 2\sqrt{x}\,dx=2\int_0^1 x^{1/2}\,dx=2\left[\frac{2}{3}x^{3/2}\right]_0^1=\frac{4}{3}.$$ 5. Compute the second integral: $$\int_1^2\big(2\sqrt{x}-(x-1)\big)dx =\int_1^2 2\sqrt{x}\,dx-\int_1^2 (x-1)\,dx.$$ Now, $$\int_1^2 2\sqrt{x}\,dx=2\left[\frac{2}{3}x^{3/2}\right]_1^2 =\frac{4}{3}(2\sqrt{2}-1).$$ Also, $$\int_1^2 (x-1)\,dx=\left[\frac{x^2}{2}-x\right]_1^2 =\left(2-2\right)-\left(\frac12-1\right)=\frac12.$$ So, $$\int_1^2\big(2\sqrt{x}-(x-1)\big)dx =\frac{4}{3}(2\sqrt{2}-1)-\frac12.

  1. Add both parts: Area=43+43(22−1)−12.\text{Area}=\frac43+\frac{4}{3}(2\sqrt{2}-1)-\frac12.Area=34​+34​(22​−1)−21​. Simplify: 43−43=0,\frac43-\frac43=0,34​−34​=0, so Area=832−12.\text{Area}=\frac{8}{3}\sqrt{2}-\frac12.Area=38​2​−21​.

  2. Compare with the options:

  • A: 832−1\frac{8}{3}\sqrt2-138​2​−1
  • B: 432+1\frac{4}{3}\sqrt2+134​2​+1
  • C: 832−12\frac{8}{3}\sqrt2-\frac1238​2​−21​
  • D: 432−12\frac{4}{3}\sqrt2-\frac1234​2​−21​

Thus the correct option is C.\boxed{\text{C}}.C​.

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