Q. 235.0( 1 Vote )

# Evaluate <span la

Answer :

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

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