For this problem, we will need to imagine the arch as a parabola on a graph. We will then need to find the equation of the parabola, and use an integral to find the area under it. Although we can place the arch almost anywhere on the graph, the easiest place to put it is in the center of the axes as shown below. The base of the arch is 2 meters wide (intersecting the x-axis at -1 and +1), and it is 3 meters high (intersecting the y-axis at 3).

We now need an equation for this parabola so that we can integrate. If you know (at least) 3 points on a parabola you can find the equation for that parabola. The 3 points we have are:

`(0,3),(-1,0),(1,0)`

And using the standard form of quadratic function `ax^2+bx+c=y`

we can substitute in our x- and y-values and find the coefficients a, b, and c.

First we have:

`a(0)^2+b(0)+c=3`

`c=3`

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` `Then we have:

`a(-1)^2+b(-1)+c=0`

`a-b+3=0`

`a-b=-3`

And last we have:

`a(1)^2+b(1)+c=0`

`a+b+3=0`

`a+b=-3`

So now we solve this small system of equations for a and b:

`a-b=-3`

`a+b=-3`

Solving gives you `a=-3 and b=0`

So now we have the following quadratic equation to represent our arch:

`y=-3x^2+3`

To find the area under a curve (here, an arch), we must take the integral of the function between the desired bounds (here, between -1 and 1). Our integral looks like this:

`A=int_-1^1-3x^2+3 dx`

Integrate each term individually using the power rule:

`A=(-3x^3)/3+3x+C`

`A=-x^3+3x+C`

Evaluate from -1 to 1:

`A=[-(1)^3+3(1)]-[-(-1)^3+3(-1)]`

`A=2-(-2)`

`A=4 m^2`

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