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Area Under the Curves question

2004 · Shift 0 · Q88
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Area Under the Curves question

2004 · Shift 0 · Q88

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area of the region bounded by the curves y=∣x−2∣,x=1,x=3y = \left| {x - 2} \right|,x = 1,x = 3y=∣x−2∣,x=1,x=3 and the xxx-axis is :
  1. A
    444
  2. B
    222
  3. C
    333
  4. D
    111
View written solutionFree

Correct answer: D

  1. Understand the region

We need the area bounded by:

  • y=∣x−2∣y = |x-2|y=∣x−2∣
  • x=1x=1x=1
  • x=3x=3x=3
  • the xxx-axis

So the required area is the area between the curve y=∣x−2∣y=|x-2|y=∣x−2∣ and the xxx-axis from x=1x=1x=1 to x=3x=3x=3.

  1. Write ∣x−2∣|x-2|∣x−2∣ in piecewise form

Since the expression changes at x=2x=2x=2,

∣x−2∣={2−x,x≤2x−2,x≥2|x-2|= \begin{cases} 2-x, & x\le 2 \\ x-2, & x\ge 2 \end{cases}∣x−2∣={2−x,x−2,​x≤2x≥2​

Thus,

  • on [1,2][1,2][1,2], y=2−xy=2-xy=2−x
  • on [2,3][2,3][2,3], y=x−2y=x-2y=x−2
  1. Set up the area integral
A=∫12(2−x) dx+∫23(x−2) dxA=\int_1^2 (2-x)\,dx + \int_2^3 (x-2)\,dxA=∫12​(2−x)dx+∫23​(x−2)dx
  1. Evaluate each part

First integral:

∫12(2−x) dx=[2x−x22]12\int_1^2 (2-x)\,dx = \left[2x-\frac{x^2}{2}\right]_1^2∫12​(2−x)dx=[2x−2x2​]12​

At x=2x=2x=2:

2(2)−222=4−2=22(2)-\frac{2^2}{2}=4-2=22(2)−222​=4−2=2

At x=1x=1x=1:

2(1)−122=2−12=322(1)-\frac{1^2}{2}=2-\frac12=\frac322(1)−212​=2−21​=23​

So,

∫12(2−x) dx=2−32=12\int_1^2 (2-x)\,dx = 2-\frac32=\frac12∫12​(2−x)dx=2−23​=21​

Second integral:

∫23(x−2) dx=[x22−2x]23\int_2^3 (x-2)\,dx = \left[\frac{x^2}{2}-2x\right]_2^3∫23​(x−2)dx=[2x2​−2x]23​

At x=3x=3x=3:

92−6=−32\frac{9}{2}-6=-\frac3229​−6=−23​

At x=2x=2x=2:

42−4=2−4=−2\frac{4}{2}-4=2-4=-224​−4=2−4=−2

So,

∫23(x−2) dx=−32−(−2)=12\int_2^3 (x-2)\,dx = -\frac32-(-2)=\frac12∫23​(x−2)dx=−23​−(−2)=21​
  1. Add the two areas
A=12+12=1A=\frac12+\frac12=1A=21​+21​=1
  1. Match with options

The area is

1\boxed{1}1​

So the correct option is D.

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