Q. 305.0( 2 Votes )

# Using integration find the area of the triangle formed by positive x-axis and tangent and normal to the circle x^{2} + y^{2} = 4 at (1, √3).

**OR**

Evaluate: as a limit of a sum. **[CBSE 2015]**

**[CBSE 2015]**

Answer :

As first we need to trace the area to be determined.As x^{2}+y^{2} = 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, x^{2} + y^{2} = 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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