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Area Under the Curves question

2020 · 8 Jan · Shift 2 · Q35
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  5. /2020 · 8 Jan · Shift 2 · Q35

Area Under the Curves question

2020 · 8 Jan · Shift 2 · Q35

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region {(x,y) ∈\in∈ R2 : x2 ≤\le≤ y ≤\le≤ 3 – 2x}, is :
  1. A
    343{{34} \over 3}334​
  2. B
    293{{29} \over 3}329​
  3. C
    313{{31} \over 3}331​
  4. D
    323{{32} \over 3}332​
View written solutionFree

Correct answer: D

  1. Identify the curves

The region is defined by x2≤y≤3−2x.x^2 \le y \le 3-2x.x2≤y≤3−2x. So the area lies between:

  • lower curve: y=x2y=x^2y=x2
  • upper curve: y=3−2xy=3-2xy=3−2x
  1. Find points of intersection

For the bounded region, the curves must intersect: x2=3−2xx^2=3-2xx2=3−2x x2+2x−3=0x^2+2x-3=0x2+2x−3=0 (x+3)(x−1)=0(x+3)(x-1)=0(x+3)(x−1)=0 So, x=−3,  1.x=-3,\;1.x=−3,1.

  1. Check which curve is above

We need area between the curves from x=−3x=-3x=−3 to x=1x=1x=1: Area=∫−31[(3−2x)−x2]dx.\text{Area}=\int_{-3}^{1}\big[(3-2x)-x^2\big]dx.Area=∫−31​[(3−2x)−x2]dx.

  1. Evaluate the integral
=\left[3x-x^2-\frac{x^3}{3}\right]_{-3}^{1}.$$ Now compute at the limits: At $x=1$, $$3(1)-1^2-\frac{1^3}{3}=3-1-\frac13=\frac53.$$ At $x=-3$, $$3(-3)-(-3)^2-\frac{(-3)^3}{3}=-9-9-\left(-9\right)=-9.$$ Therefore, $$\text{Area}=\frac53-(-9)=\frac53+9=\frac{5+27}{3}=\frac{32}{3}.$$ 5. **Match with options** $$\boxed{\frac{32}{3}}$$ So the correct option is **D**.
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