(i) We place the parabola so that the left side of the arch passes through `(0, 0)` and the right side will pass through `(2, 0)`, since the arch is 2 m wide at the bottom. The vertex of the arch is at `(1, 3)`.

**General form of a parabola:** `y =
ax^2+
bx +
c`

At `x = 0`, `y = 0` and on substituting, we get `0 = 0 + 0 + c`. So `c = 0`.

At `x = 1`, `y = 3` and on substituting, we get `3 = a + b`.

At `x = 2`, `y = 0` and on substituting, we get `0 = 4a + 2b`.

This gives the simultaneous equations

`a + b = 3`

`2a + b = 0`

Subtracting the 1st line from the 2nd gives `a = -3`.

And so `b = 6`.

So the required parabola is `y =
-3x^2+
6x`** ,** with `x`

The process of finding the equation is called
**modeling**. It is a very important skill in science and
engineering.

(ii) Now for the area:

`int_0^2(-3x^2+6x)dx =[-x^3+3x^2]_0^2`

`=[-(2^3)+3(2)^2]-[0+0]`

`=[-8+12]`

`=4\ "m"^2`

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