Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.

OR

Using integration, find the area of the triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C(8, 4).

Given, A circle has equation 4x² + 4y² = 9 and a parabola x² = 4y.

To Find: Find the area of a circle.

Explanation: We have a circle equation 4x² + 4y² = 9 and parabola x² = 4y

We can write the given equation as

- - - - (i)

And, x² = 4y - - - - (ii)

Now, Put the value of x² 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