Problem 113 - Calculus Volume 1

Question

In the following exercises, find the average value \(f_{\text { ave }}\) of \(f\) between \(a\) and \(b,\) and find a point \(c,\) where \(f(c)=f_{\text { ave. }}\) \(f(x)=(3-|x|), a=-3, b=3\)

Step-by-Step Solution

Answer

The average value is \( \frac{3}{2} \) and \( c = \pm \frac{3}{2} \).

Step 1: Calculate Average Value Formula

The average value of a function \( f(x) \) over an interval \([a, b]\) is given by the formula: \[ f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] In our exercise, \( a = -3 \) and \( b = 3 \), so we need to compute \( \frac{1}{6} \int_{-3}^{3} (3 - |x|) \, dx \).

Step 2: Integrate the Function

First, note \(|x|\) is piecewise: \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\). Thus, split the integral at 0: \[ \int_{-3}^{3} (3 - |x|) \, dx = \int_{-3}^{0} (3 + x) \, dx + \int_{0}^{3} (3 - x) \, dx \] Let’s compute these integrals separately.

Step 3: Compute the First Integral

The first integral becomes: \[ \int_{-3}^{0} (3 + x) \, dx = \left[ 3x + \frac{x^2}{2} \right]_{-3}^{0} = \left(3(0) + \frac{0^2}{2}\right) - \left(3(-3) + \frac{(-3)^2}{2}\right) = 0 - (-9 + 4.5) = 4.5 \]

Step 4: Compute the Second Integral

The second integral becomes: \[ \int_{0}^{3} (3-x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{3} = \left(3(3) - \frac{3^2}{2}\right) - \left(3(0) - \frac{0^2}{2}\right) = 9 - 4.5 = 4.5 \]

Step 5: Combine Integrals to Calculate Average Value

Add the results of the two integrals: \(4.5 + 4.5 = 9\). Now use the average value formula: \[ f_{\text{ave}} = \frac{1}{6} \times 9 = \frac{3}{2} \]

Step 6: Find the Point where \(f(c)=f_{\text{ave}}\)

We need to find \(c\) such that \(f(c) = \frac{3}{2}\). For \(f(x) = 3 - |x|\), solve \(3 - |c| = \frac{3}{2}\): \[ |c| = 3 - \frac{3}{2} = \frac{3}{2} \] Thus, \(c = \pm \frac{3}{2}\).

Key Concepts

Definite Integral

The concept of the definite integral is a fundamental aspect of calculus, capturing the notion of the accumulation of quantities over an interval. When we compute a definite integral, we essentially sum up small quantities of interest between two bounds, labeled as \(a\) and \(b\).
The definite integral of a function \(f(x)\) from \(a\) to \(b\) is represented as \(\int_{a}^{b} f(x) \, dx\). This integral, geometrically, is the net area under the curve \(f(x)\) over the interval \([a, b]\), taking into account the signs of the areas.
For this specific exercise, we used the definite integral to find the average value of the function \(f(x) = 3 - |x|\) over the interval \([-3, 3]\). By evaluating the integral, we determine the total accumulation of the function values and relate this to the interval length to find the average.

Piecewise Functions

Piecewise functions are a powerful tool in mathematics, allowing us to define a function that acts differently in various parts of its domain. Such functions are divided into sections, each defined by its own expression.
In this exercise, the function \(f(x) = 3 - |x|\) is piecewise due to the nature of the absolute value. The absolute value function \(|x|\) leads to two cases:

• \(|x| = x\) when \(x \geq 0\)
• \(|x| = -x\) when \(x < 0\)

Thus, to evaluate the definite integral of \(3 - |x|\) from \(-3\) to \(3\), the function was split into two parts:

• \(3 + x\) for \(x < 0\) (interval \([-3, 0]\))
• \(3 - x\) for \(x \geq 0\) (interval \([0, 3]\))

This segmentation simplifies the integration process, ensuring accuracy by respecting the function’s behavior over different intervals.

Absolute Value

Understanding absolute value is crucial for handling piecewise functions. The absolute value of a number \(x\), denoted \(|x|\), represents the distance of \(x\) from zero on the number line, always yielding a non-negative result.
The formula is straightforward:

• \(|x| = x\), if \(x \geq 0\)
• \(|x| = -x\), if \(x < 0\)

The exercise utilized the absolute value to modify the behavior of the function \(f(x) = 3 - |x|\). It required handling different expressions for \(|x|\) based on whether \(x\) was positive or negative. This ensured that during the integration and analysis of \(f(x)\), the correct mathematical statements were applied according to the sign of \(x\). As a result, the piecewise nature accounted for the absolute values, facilitating accurate integration over the specified intervals.

Integration Techniques

Integration techniques are methods used to solve integrals, enabling us to find areas under curves or solve complex problems. For some functions, direct integration is possible using basic rules. However, when handling more complicated functions, like those involving piecewise sections, more specific techniques are necessary.
In this exercise, integration was split into two distinct parts due to the piecewise nature of \(f(x) = 3 - |x|\):

• For \(3 + x\) in \([-3, 0]\), the integration involved adding \(3\) and \(x\): \(\int (3 + x) \, dx\), resulting in \(3x + \frac{x^2}{2}\).
• For \(3 - x\) in \([0, 3]\), the integration required subtracting \(x\) from \(3\): \(\int (3 - x) \, dx\), resulting in \(3x - \frac{x^2}{2}\).

Each separate integral was computed over its respective domain, ensuring the correct calculation and interpretation of areas. Combining these results allowed for the determination of the average function value over the entire interval.