ExamHabit
Practise free

JEE Main 2019 · 9 Apr · Shift I

JEE Main 2019 Mathematics question on Definite Integrals and Areas

The area (in sq units) of the region A={(x,y):x2≤y≤x+2}A = \{(x, y) : x^2 \leq y \leq x + 2\} is
  1. A.
    136\frac{13}{6}
  2. B.
    92\frac{9}{2}
  3. C.
    316\frac{31}{6}
  4. D.
    103\frac{10}{3}
Show answer and solution

Correct option: B

Solution

The curves intersect where x² = x + 2, giving x² - x - 2 = 0, so x = -1 and x = 2. The area is the integral from -1 to 2 of (x + 2 - x²) dx = [x²/2 + 2x - x³/3] from -1 to 2 = (2 + 4 - 8/3) - (1/2 - 2 + 1/3) = (18/3 - 8/3) - (-9/6) = 10/3 + 3/2 = 20/6 + 9/6 = 29/6... let me recompute: at x=2: 4/2 + 4 - 8/3 = 2 + 4 - 8/3 = 6 - 8/3 = 10/3. At x=-1: 1/2 - 2 + 1/3 = 3/6 - 12/6 + 2/6 = -7/6. Area = 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2. The parabola y = x² lies below y = x + 2 between x = -1 and x = 2, so integrating (upper - lower) gives 9/2.

[JEE Main 2019 · 9 Apr · Shift I]

Practise Definite Integrals and Areas the way it is asked

A timed set of real JEE Main Definite Integrals and Areas questions, graded as the exam grades. No sign-up.

Start a Definite Integrals and Areas test

More: all JEE Main Definite Integrals and Areas questions · JEE Main 2019 · Mathematics chapter list