Q. 21

# Using integration

As first we need to trace the area to be determined.As x2+y2 = 4 represents a circle whose centre is at (0,0) and radius = 2 cm. The rough sketch is shown below:

As normal at (1,√3) passes through origin too because in a circle a normal always passes through centre of the circle.

Equation of normal is y = √3 x

Similarly equation of tangent can be written using one point and slope form.

As, x2 + y2 = 4

Differentiating w.r.t x:

We need to determine the area enclosed i.e. area(region ABC).

Area(region ABC) = area(region ABD) + area(region BDC)

Area = area under curve y = √3x + area under

Required area =

Area =

Area =

Area =

Required area =

OR

We know that a definite integral can be evaluated as a limit of a sum as-

Where h =

As we have to find:

Let I = and on comparing I with the formula we can say that a = 1 and b = 3.

I =

I = 2

I = 2

I = 2

Each bracket contains ‘n’ terms.

I = 2

I = 2

I = 2

Using formula for sum of first n natural numbers; sum of squares of first n natural numbers and sun of n terms of a GP we get:

I = 2

I = 2

As h =

I = 2

I = 2

Using algebra of limits:

I = 2

Use the formula:

I = 2

I =

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