The area of the region {(x,y):x²+4x+2≤ y≤ |x+2|} is equal to

Solution

Maths sir · 4 steps

Shift to t = x + 2 and the whole region becomes symmetric about t = 0, so find one half and double it.

1. Shift the variable. Complete the square: x² + 4x + 2 = (x+2)² - 2. Put t = x + 2. The region becomes t² - 2 ≤ y ≤ |t|. A shift does not change area.
2. Use symmetry. Both t² - 2 and |t| are even in t, so the region is symmetric about t = 0. For t ≥ 0 the upper curve is y = t. It meets the parabola where t² - 2 = t, that is (t-2)(t+1) = 0, so at t = 2.

   Area = 2∫₀²(t - t² + 2)dt = 2[t²/2 - t³/3 + 2t]₀²

3. Evaluate. At t = 2 the bracket is 2 - 8/3 + 4 = 10/3, and at t = 0 it is 0. So one half has area 10/3.
4. So. The total area is 2 × 10/3 = 20/3, which is option (c).

The idea to keep. When a modulus and a parabola share the same centre, shift the origin to that centre. The region usually turns symmetric, and you integrate only one side.

Answer from the NTA final answer key. Solution by PrepWiser.