The area of the region enclosed by the curves y=x²-4x+4 and y²=16-8x is :

Solution

Maths sir · 4 steps

Spot that y=(x-2)², find where the two parabolas meet, then integrate top minus bottom from 0 to 2.

1. Recognise the curves. The first curve is y=(x-2)², an upward parabola with vertex (2,0). The second is y²=8(2-x), a parabola opening to the left, also with vertex (2,0). Since y=(x-2)² ≥ 0, it can only meet the upper half y=√(16-8x).
2. Find the meeting points. Put y=(x-2)² into y²=-8(x-2): (x-2)⁴=-8(x-2). So x-2=0 or (x-2)³=-8, which gives x=2 or x=0. The points are (2,0) and (0,4).
3. Set up the integral. At x=1 the left-opening parabola gives y=√8≈ 2.83 and the other gives y=1, so √(16-8x) is on top throughout 0 ≤ x ≤ 2.

   A=∫₀² √(16-8x) dx-∫₀² (x-2)² dx=16/3-8/3

4. So. With u=16-8x, the first integral is 1/8∫₀¹⁶√u du=1/8×2/3× 64=16/3, and ∫₀² (x-2)² dx=8/3. The area is 8/3, option (a).

The idea to keep. Before integrating, rewrite each curve in vertex form. Shared vertices and easy intersection points often appear, and then area is just top curve minus bottom curve between those points.

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