Question

टाइम टेकन टो कॉन्टैक्ट्स फ्रॉम वन टू हंड्रेड​

Answer 1

Answer:

The area encoded by [x](x - 1) ≤ y ≤ 2√x from x = 0 to 2 where [x] is the greatest integer less than or equal to x, is equal to ....

solution : [x](x - 1) ≤ y ≤ 2√x from x = 0 to 2

for x ≥ 0 ⇒y = 2√2x ⇒y² = 4x [ a parabolic graph ]

draw y² = 4x in interval x ≥ 0, diagram is shown in figure.

now let's solve the piece wise function I mean great integer function.

y = [x](x - 1)

case 1 : 0 ≤ x < 1 , y = 0

case 2 : 1 ≤ x < 2 , y = 1(x - 1) = x - 1

case 3 : x = 2 , y = 2(2 - 1) = 2

now draw y = [x](x - 1) with y² = 4x, you will get the enclosed area (shaded region) as shown in figure.

now area enclosed by the curves , A = 

= 4/3[x³/²]²₀ - 1/2

= 4/3 × (2√2) - 1/2

= 8√2/3 - 1/2

Therefore the area enclosed by th