Evaluate as limit of sums.

OR

Using integration, find the area of the following region:

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

We have,

Let f(x) = 3x²+ 2x …(i)

Here,

a = 1

b = 3

nh = b – a = 3 – 1 = 2 …(*)

We know limit of sums is given as,

Put a = 1 as deduced above.

Put x = 1 in (i),

f(1) = 3(1)² + 2(1)

Put x = 1 + h in (i),

f(1 + h) = 3(1 + h)² + 2(1 + h)

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

From (*), nh = 2

=10+8+16

OR

We have two equations:

Let us take equation of ellipse and find y.

Taking square root on both sides,

Let it be y₁, such that

…(i)

Let us take equation of straight line and find y.

Let it be y₂, such that

…(ii)

Put in .

⇒ √(9 – x²) = 3 – x

Squaring on both sides, we get

⇒ [√(9 – x²)]² = (3 – x)²

⇒ 9 – x² = 9 + x² – 6x

⇒ x² + x² + 9 – 9 – 6x = 0

⇒ 2x² – 6x = 0

⇒ 2x(x – 3) = 0

⇒ 2x = 0 or (x – 3) = 0

⇒ x = 0 or x = 3

Put x = 0 in .

⇒ y = 2

Put x = 3 in .

⇒ y = 0

One point of intersection is (0, 2) and the other point of intersection is (3, 0).

So, the area of the region is given by

Putting the values of y₁ and y₂ from (i) and (ii) in the above equation, we get