Evaluate as limit of sums.

OR

Using integration find the area of the following region:

{(x, y): |x – 1| ≤ y ≤ √(5 – x²)}

Know that, limit of sums is the sum of limits of the function.

We have,

Let f(x) = x²– x …(i)

Here,

a = 1

b = 4

nh = b – a = 4 – 1 = 3 …(*)

We know limit of sums is given as,

Put a = 1 as deduced above.

Put x = 1 in (i),

f(1) = 1 – 1

Put x = 1 + h in (i),

f(1 + h) = (1 + h)² – (1 + h)

Repeat the same for other values of x. We get,

∵, nh = 3

OR

We have two equations:

One of them is the equation of curve, say y₁.

y = √(5 – x²)

Squaring on both sides,

y² = (√(5 – x²))²

⇒ y² = 5 – x²

⇒ x² + y² = 5

It is a circle.

The other equation is of a line,

y = |x – 1|

Take y = x – 1 and y = 1 – x …(ii)

Let us find intersection point of these two figures by substituting y = x – 1 in y = √(5 – x²). We get,

x – 1 = √(5 – x²)

Squaring both sides,

(x – 1)² = [√(5 – x²)]²

⇒ x² + 1 – 2x = 5 – x²

⇒ x² + x² – 2x + 1 – 5 = 0

⇒ 2x² – 2x – 4 = 0

⇒ x² – x – 2 = 0

⇒ x² – 2x + x – 2 = 0

⇒ x(x – 2) + (x – 2) = 0

⇒ (x + 1)(x – 2) = 0

⇒ (x + 1) = 0 or (x – 2) = 0

⇒ x = -1 or x = 2

Put x = -1 in y = x – 1,

y = -1 – 1

⇒ y = -2

Point of intersection of circle x² + y² = 5 and line y = x – 1 is (-1, -2).

Put x = 2 in y = x – 1,

y = 2 – 1

⇒ y = 1

Point of intersection between x² + y² = 5 and line y = 1 – x is (2, 1).

So, the area of the region is given by