Refer the attached image with calculations.

The equation of an arch is a quadratic function.

Let us say the equation of the arch is;

`y = ax^2+bx+c`

Now we draw the arch so that the left most side of the arch is passes through (0,0) as in the attachment.

According to the graph of the arch and due to symmetry;

At x = 0 then y = 0 -----(1)

At x = 1 then y = 3 -------(2)

At x = 2 then y = 0 -------(3)

`y = ax^2+bx+c`

From (1);

`0 = 0+0+c`

`c = 0`

From (2)

`3 = axx1+bxx1+c`

`3 = a+b+0`

`3 = a+b -----(4)`

From (3)

`0 = axx(2)^2+bxx2+c`

`b = -2a ----(5)`

Solving (4) and (5) yields;

`a = -3`

`b = 6`

`y = -3x^2+6x`

Area of the arch is given by integration.

Area under arch

`= int_0^2ydx`

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

`= [-x^3+3x^2]_0^2`

`= (-8+12)`

`= 4`

**So the area under the curve is `4m^2` .**

**Further Reading**

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