Answer :

**Given,** A circle has equation 4x^{2} + 4y^{2} = 9 and a parabola x^{2} = 4y.

**To Find:** Find the area of a circle.

**Explanation:** We have a circle equation 4x^{2} + 4y^{2} = 9 and parabola x^{2} = 4y

We can write the given equation as

- - - - (i)

And, x^{2} = 4y - - - - (ii)

Now, Put the value of x^{2} in equation (i)

2y(2y + 9) - 1(2y + 9) = 0

(2y + 9)(2y - 1) = 0

Here, We neglect the negative value

Substitute the value of y in equation (ii)

Since,

Area of required region = 2 (Area in the first quadrant)

**Hence, This is the required Area of given equation.**

**OR**

**Given****,** A triangle ABC , whose vertices are A(4, 1) , B(6, 6) and C(8, 4)

**To Find:** Find the area of triangle ABC using integration.

**Explanation:** We have three vertices of triangle A(4, 1) , B(6, 6) and C(8, 4).

Now, The equation of line AB is,

2y - 2 = 5x - 20

2y = 5x - 18

Now, The equation of BC is

y - 6 = - 1(x - 6)

BC = y = 12 - x

Similarly,

The equation of AC is

4y - 4 = 3x - 12

Area of ∆ABC = Area under AB + Area under BC – Area under AC

square units

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