An arch is 

675675 

ft high and has a 

560560​-ft 

base. It can be modeled by the parabola 

yequals=675 left bracket 1 minus left parenthesis StartFraction x Over 280 EndFraction right parenthesis squared right bracket6751−x2802. 

Find the average height of the arch above the ground.

-400-2000200400100200300400500600700xy

A coordinate system has a horizontal x-axis labeled from negative 400 to 400 in increments of 100 and a vertical y-axis labeled from 0 to 700 in increments of 50. A parabola opens downward from (0, 675) and passes through (negative 200, 331) and (200, 331). All coordinates are approximate.

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Part 1

The average height of the arch is 

enter your response here 

ft above the ground.

To find the average height of the arch above the ground, we need to calculate the average value of the given function over the interval representing the base of the arch. The function representing the height of the arch is given by:

\[ y = 675 \left( 1 - \left( \frac{x}{280} \right)^2 \right) \]

The base of the arch extends from \( x = -280 \) to \( x = 280 \).

The formula for the average value of a function \( f(x) \) over the interval \([a, b]\) is:

\[ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx \]

In this case, \( f(x) = 675 \left( 1 - \left( \frac{x}{280} \right)^2 \right) \), \( a = -280 \), and \( b = 280 \). Therefore, we need to evaluate:

\[ \text{Average height} = \frac{1}{280 - (-280)} \int_{-280}^{280} 675 \left( 1 - \left( \frac{x}{280} \right)^2 \right) \, dx \]

Simplifying the interval:

\[ \text{Average height} = \frac{1}{560} \int_{-280}^{280} 675 \left( 1 - \left( \frac{x}{280} \right)^2 \right) \, dx \]

We can split the integral into two parts:

\[ \text{Average height} = \frac{675}{560} \left( \int_{-280}^{280} 1 \, dx - \int_{-280}^{280} \left( \frac{x}{280} \right)^2 \, dx \right) \]

First, evaluate the integral of 1 over the interval:

\[ \int_{-280}^{280} 1 \, dx = x \Big|_{-280}^{280} = 280 - (-280) = 560 \]

Next, evaluate the integral of \( \left( \frac{x}{280} \right)^2 \) over the interval:

\[ \int_{-280}^{280} \left( \frac{x}{280} \right)^2 \, dx = \frac{1}{280^2} \int_{-280}^{280} x^2 \, dx \]

Using the property of symmetry for even functions over symmetric intervals, we have:

\[ \int_{-280}^{280} x^2 \, dx = 2 \int_{0}^{280} x^2 \, dx \]

Now, compute the integral:

\[ \int_{0}^{280} x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{280} = \frac{280^3}{3} - \frac{0^3}{3} = \frac{280^3}{3} \]

Thus,

\[ \int_{-280}^{280} x^2 \, dx = 2 \cdot \frac{280^3}{3} = \frac{2 \cdot 280^3}{3} \]

Then,

\[ \int_{-280}^{280} \left( \frac{x}{280} \right)^2 \, dx = \frac{1}{280^2} \cdot \frac{2 \cdot 280^3}{3} = \frac{2 \cdot 280}{3} = \frac{560}{3} \]

Finally, combine the results:

\[ \text{Average height} = \frac{675}{560} \left( 560 - \frac{560}{3} \right) \]
\[ = \frac{675}{560} \left( \frac{1680 - 560}{3} \right) \]
\[ = \frac{675}{560} \cdot \frac{1120}{3} \]
\[ = \frac{675 \cdot 1120}{560 \cdot 3} \]
\[ = \frac{675 \cdot 2}{3} \]
\[ = 450 \]

Therefore, the average height of the arch above the ground is 450 feet.

Would you like any further details or have any questions?

Here are some additional questions you might consider:
1. How can you use symmetry properties to simplify integrals?
2. What are the general properties of parabolas and their equations?
3. How do you interpret the result of an average value in the context of geometry?
4. What techniques can be used to evaluate definite integrals more efficiently?
5. How does the average value theorem apply to real-world scenarios?

**Tip:** When dealing with symmetric functions over symmetric intervals, remember that even functions (like \(x^2\)) have simpler integral evaluations due to their symmetry.

Explore how to calculate the average height of a parabolic arch using integration and the average value theorem. This mathematical problem involves finding the average height of an arch modeled by a parabola, with detailed steps and calculations. Suitable for advanced high school students and beyond.

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