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Chapter 6: Problem 7

Let \(f(x)=2-x^{2}\). Recall that the average value of any continuous function \(f\) on an interval \([a, b]\) is given by \(\frac{1}{b-a} \int_{a}^{b} f(x) d x\) a. Find the average value of \(f(x)=2-x^{2}\) on the interval \([0, \sqrt{2}]\). Call this value \(r\). b. Sketch a graph of \(y=f(x)\) and \(y=r\). Find their intersection point(s). c. Show that on the interval \([0, \sqrt{2}],\) the amount of area that lies below \(y=f(x)\) and above \(y=r\) is equal to the amount of area that lies below \(y=r\) and above \(y=f(x)\). d. Will the result of (c) be true for any continuous function and its average value on any interval? Why?

Short Answer

Expert verified
The average value is \(\frac{4}{3}\). Intersection points are \(x = \sqrt{\frac{2}{3}}\) and \(x = -\sqrt{\frac{2}{3}}\). The areas above and below \(y = r\) are equal by symmetry.

Step by step solution

01

Set up the integral for the average value

To find the average value of the function on the given interval, use the formula for the average value of a continuous function. For the given function on the interval \([0, \sqrt{2}]\), set up the integral as follows: \(\frac{1}{\sqrt{2} - 0} \int_{0}^{\sqrt{2}} (2 - x^{2}) dx\).
02

Compute the definite integral

Next, evaluate the definite integral \(\int_{0}^{\sqrt{2}} (2 - x^{2}) dx\). Split the integral and solve: \[\int_{0}^{\sqrt{2}} 2 dx - \int_{0}^{\sqrt{2}} x^{2} dx\]. Evaluate these separately:
03

Solve the integrals

Evaluate each part of the integral: \[\int_{0}^{\sqrt{2}} 2 dx = 2x \bigg|_{0}^{\sqrt{2}} = 2(\sqrt{2}) - 2(0) = 2\sqrt{2} \]. For the second integral: \[\int_{0}^{\sqrt{2}} x^{2} dx = \frac{x^{3}}{3} \bigg|_{0}^{\sqrt{2}} = \frac{(\sqrt{2})^{3}}{3} - \frac{(0)^{3}}{3} = \frac{2\sqrt{2}}{3} \]. Combine these results to get: \[\int_{0}^{\sqrt{2}} (2 - x^{2}) dx = 2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{6\sqrt{2}}{3} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3} \].
04

Find the average value

Divide the result of the integral by the length of the interval \([0, \sqrt{2}]\): \[\frac{1}{\sqrt{2} - 0} \frac{4\sqrt{2}}{3} = \frac{4\sqrt{2}}{3\sqrt{2}} = \frac{4}{3} \]. Thus, the average value \(r\) is \(\frac{4}{3}\).
05

Sketch the graph

Graph the function \(y = f(x) = 2 - x^{2}\) and the horizontal line \(y = r = \frac{4}{3}\). The intersection points can be found by solving \(2 - x^{2} = \frac{4}{3}\).
06

Find intersection points

Solve \(2 - x^{2} = \frac{4}{3}\): \[\begin{aligned} 2 - \frac{4}{3} &= x^{2} \Rightarrow \frac{6}{3} - \frac{4}{3} &= x^{2} \Rightarrow \frac{2}{3} &= x^{2} \Rightarrow x &= \pm\sqrt{\frac{2}{3}} \end{aligned} \]. So, the intersection points are \(x = \sqrt{\frac{2}{3}}\) and \(x = -\sqrt{\frac{2}{3}}\).
07

Show equal areas

On the interval \([0, \sqrt{2}]\), the area above and below \(y = r\) and \(y = f(x)\) are equal by symmetry of the parabola around its vertex \(x = 0\). This implies that the areas above and below the average value are balanced.
08

Generalize for any continuous function

The result in Step 7 will be true for any continuous function on any interval because the average value \(r \) is designed to balance the total area above and below it. By definition, the integral of the deviations above and below must sum to zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integrals
A definite integral, represented as \(\int_{a}^{b} f(x) \ dx\), is a way to calculate the accumulation of quantities, such as areas under a curve. For a given function, it calculates the net signed area between the curve of the function and the x-axis over an interval [a, b].

Key points about definite integrals:
  • The integral provides a way to add up infinitely many small quantities, which can be areas, lengths, or other measurements.
  • Splitting an integral into smaller parts and evaluating each part separately can simplify the calculation process.
  • When you evaluate a definite integral, you compute the differences of the antiderivative at the bounds.
In our example, we calculated \(\int_{0}^{\sqrt{2}} (2 - x^{2}) \ dx\) to find the total area under the curve of \(\ f(x)=2-x^{2} \) from \(\ x = 0 \) to \(\ x = \ \sqrt{2} \).

This integral was split and each part was solved separately to eventually combine them and obtain the final result.
Continuous Functions
A continuous function is one that can be drawn without lifting your pencil off the paper. This means there are no abrupt jumps or breaks.
Key points about continuous functions:
  • A function \( f(x) \) is continuous over an interval if it is continuous at every point within that interval.
  • Calculations like finding the area under the curve using integrals rely on the function being continuous so that every tiny segment of the curve can be accurately summed up.
  • In calculus, continuous functions are essential because they ensure that the integrals yield meaningful results.
In our given exercise, \(\ f(x) = 2 - x^{2} \) is continuous on \(\ [0, \ \sqrt{2}] \).

This continuity is what allows us to use integral calculus to find the average value effectively. Without continuity, the integral might not account for sudden changes or gaps.
Area Under a Curve
Understanding the area under a curve helps in various contexts like physics for computing work done by a force or in probability to find distributions.
Key points about the area under a curve:
  • The area under the curve \( f(x) \) from \( a \) to \( b \) represents the accumulation of the values of \( f(x) \) over that interval.
  • The definite integral of \( f(x) \) over \( [a, b] \) gives this area, accounting for whether the curve lies above or below the x-axis.
  • Positive areas (above the x-axis) and negative areas (below the x-axis) are combined, giving the net area.
In the exercise, we demonstrate that the area under the curve of \( y = f(x) = 2 - x^{2} \) above the average value \(\frac{4}{3}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\
  • of thearea**

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    Most popular questions from this chapter

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