Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Area Under Curves

NTA Abhyas 2020 · Mathematics · Area Under Curves

The area bounded by the curves y = (x - 1) ^ 2 , y = (x + 1) ^ 2...

The area bounded by the curves y=(x−1)2,y=(x+1)2\text{y} = {\left(\text{x} - 1\right)}^{2} \text{,} \text{y} = {\left(\text{x} + 1\right)}^{2} and y=14\text{y} = \dfrac{1}{4} isThe area bounded by the curves y = (x - 1) ^ 2 , y = (x + 1) ^ 2 and y = (1)/(4) is

Options

  1. A13sq unit\dfrac{1}{3} \text{sq unit}
  2. B23sq unit\dfrac{2}{3} \text{sq unit}
  3. C14sq unit\dfrac{1}{4} \text{sq unit}
  4. D15sq unit\dfrac{1}{5} \text{sq unit}

Correct answer

A. 13sq unit\dfrac{1}{3} \text{sq unit}

Step-by-step solution

The curves y=(x−1)2,y=(x+1)2\text{y} = {\left(\text{x} - 1\right)}^{2} \text{,} \text{y} = {\left(\text{x} + 1\right)}^{2} and y=14\text{y} = \dfrac{1}{4} are shown asThe area bounded by the curves y = (x - 1) ^ 2 , y = (x + 1) ^ 2 and y = (1)/(4) is

where point of intersection are

(x−1)2=14{\left(\text{x} - 1\right)}^{2} = \dfrac{1}{4}

⇒⇒ x=12\text{x} = \dfrac{1}{2}

and (x+1)2=14⇒x=−12{\left(\text{x} + 1\right)}^{2} = \dfrac{1}{4} ⇒ x = - \dfrac{1}{2}

∴∴ Q(12,14)\text{Q} \left(\dfrac{1}{2} \text{,} \dfrac{1}{4}\right) and R(−12,14)\text{R} \left(- \dfrac{1}{2} \text{,} \dfrac{1}{4}\right)

∴∴ Required area =2∫01/2[(x−1)2−14]dx= 2 {∫}_{0}^{1 / 2} \left[{\left(\text{x} - 1\right)}^{2} - \dfrac{1}{4}\right] \text{dx}

=2[(x−1)33−14x]01/2= 2 {\left[\dfrac{{\left(\text{x} - 1\right)}^{3}}{3} - \dfrac{1}{4} \text{x}\right]}_{0}^{1 / 2}

=2[−18⋅3−18−(−13−0)]=824=13sq unit= 2 \left[- \dfrac{1}{8 \cdot 3} - \dfrac{1}{8} - \left(- \dfrac{1}{3} - 0\right)\right] = \dfrac{8}{2 4} = \dfrac{1}{3} \text{sq unit}
Practice Area Under Curves on Quantrex Academy →

More from Area Under Curves

Passage: Consider the curve C_1 given by y = e^ -x for x ∈ [0, 10π] , and the curve C_2… JEE Advanced 2026Passage: Consider the ellipses given by x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1 . Question: If… JEE Advanced 2026The area of the region \ (x, y) : x^2 - 8x ≤ y ≤ -x\ is : JEE Main 2026The area of the region \ (x, y) : 0 ≤ y ≤ 6 - x, y^2 ≥ 4x - 3, x ≥ 0\ is: JEE Main 2026The area of the region R = \ (x, y): xy ≤ 27, 1 ≤ y ≤ x^2\ is equal to: JEE Main 2026The area of the region bounded by the curves x+3y^2=0 and x+4y^2=1 is equal to: JEE Main 2026The area of the region \ (x, y): y ≤ π - |x|, y ≤ |x x|, y ≥ 0\ is: JEE Main 2026If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y = 24 is A, then 3(A +… JEE Main 2026Let P_ 1 : y=4 x^ 2 and P_ 2 : y=x^ 2 +27 be two parabolas. If the area of the bounded… JEE Main 2026The area of the region R=\ (x, y): x y ≤ 8,1 ≤ y ≤ x^ 2 , x ≥ 0\ is JEE Main 2026Let f(α) denote the area of the region in the first quadrant bounded by x=0, x=1, y^ 2 =x… JEE Main 2026Let A_ 1 be the bounded area enclosed by the curves y=x^ 2 +2, x+y=8 and y -axis that… JEE Main 2026The area of the region enclosed between the circles x^ 2 +y^ 2 =4 and x^ 2 +(y-2)^ 2 =4… JEE Main 2026Let the area of the region bounded by the curve y= \ x, x\ , lines x=0, x=(3 π)/(2) , and… JEE Main 2026The area of the region A=\ (x, y): 4 x^ 2 +y^ 2 8. and .y^ 2 4 x\ is: JEE Main 2026Let the line x=-1 divide the area of the region \ (x, y): 1+x^ 2 ≤ y ≤ 3-x\ in the ratio… JEE Main 2026 Full Area Under Curves list Area Under Curves JEE Maths PYQsJEE Mathematics PYQs All NTA Abhyas JEE Main PYQs