Problem 110 - 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 \(f(c)=f_{\text { ave. }}\) \(f(x)=x^{2}, a=-1, b=1\)

Step-by-Step Solution

Answer

The average value is \(\frac{1}{3}\) and the points \(c\) are \(\pm \sqrt{\frac{1}{3}}\).

Step 1: Understand the Formula for Average Value

The average value of a function \( f(x) \) over the interval \([a, b]\) is given by the formula \( f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). This formula will be used to calculate the average value, \( f_{\text{ave}} \), of \( f(x) = x^2 \) over the interval \([-1, 1]\).

Step 2: Set Up the Integral

The function \( f(x) = x^2 \) is given, and the interval is \([-1, 1]\). The formula becomes \( f_{\text{ave}} = \frac{1}{1 - (-1)} \int_{-1}^{1} x^2 \, dx \). This simplifies to \( \frac{1}{2} \int_{-1}^{1} x^2 \, dx \).

Step 3: Evaluate the Integral

Find \( \int_{-1}^{1} x^2 \, dx \). The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). Evaluate this from \(-1\) to \(1\):\[ \left[ \frac{x^3}{3} \right]_{-1}^{1} = \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \]

Step 4: Calculate the Average Value

Substitute the evaluated integral back into the formula for the average value:\[ f_{\text{ave}} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}. \] This is the average value of the function \( f(x) = x^2 \) over the interval \([-1, 1]\).

Step 5: Find the Point \( c \)

We need to find a point \( c \) such that \( f(c) = f_{\text{ave}} = \frac{1}{3} \). Substitute \( f(c) = c^2 \) to get:\[ c^2 = \frac{1}{3}. \]Solving for \( c \) gives: \( c = \pm \sqrt{\frac{1}{3}}. \)Thus, the points \( c \) are \( \sqrt{\frac{1}{3}} \) and \( -\sqrt{\frac{1}{3}} \).

Key Concepts

Integral Calculus

Integral calculus deals with the concept of integration, which is the process of finding the integral of a function. This contrasts with differential calculus, which focuses on derivatives. Integration is essentially the reverse operation to differentiation.

In simpler terms, while differentiation breaks a function into smaller parts to find its slope, integration glues these pieces back together to find the area under the curve. Therefore, it's often related to solving real-world problems involving areas, volumes, and cumulative totals.

• A familiar analogy for integration is summing up small slices of a shape to find its total size.
• The symbol used for integration is an "elongated S" (∫), showing this sum concept visually.

Think of integral calculus as the toolkit we use when we need to "accumulate" quantities or determine how much total area lies underneath a curve over a specific interval.

Definite Integral

A definite integral is a specific type of integral that calculates the net area under a curve between two points. The two points, known as the limits of integration, are denoted by the lower and upper bounds in the integral symbol.

For example, when finding the average value of a function over a specific range, as in the exercise, we use:\[ \int_{a}^{b} f(x) \, dx\]

• This finds the total area between the function and the x-axis from point \(a\) to point \(b\).
• Often, it involves not just adding up areas above the x-axis but also subtracting areas below it.

The definite integral solves the practical need to calculate totals within boundary conditions, making it vital for numerous applications in science and engineering.

Antiderivative

An antiderivative of a function is another function whose derivative matches the original function. Identifying antiderivatives is a key component in solving integrals, specifically indefinite integrals which don't have any bounds.

In the given problem, finding the antiderivative of \(f(x) = x^2\) is crucial in computing the integral. The antiderivative here is \(\frac{x^3}{3}\). This step allows us to apply the limits and find the definite integral's result easily.

• Every function has many antiderivatives, differing only by a constant called the constant of integration \(C\).
• For example, the antiderivative of \(x^2\) is \(\frac{x^3}{3} + C\).

Calculating an antiderivative is like unraveling a complex function back into a simpler form, where it originated before going through differentiation.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concepts of differentiation and integration, essentially marrying the two mathematical operations. It states that differentiation and integration are, in a sense, inverse operations.

This theorem has two main parts:

• First Part: It states that if you have a function that's continuous on an interval and you form its antiderivative, you can use this antiderivative to calculate a definite integral by evaluating the antiderivative at the boundary points.
• Second Part: It tells us the cumulative sum of rates (derivatives) of a function gives back the original function, which answers how much change there has been over a period from \(a\) to \(b\).

In practical terms, it allows you to evaluate integrals easily given an antiderivative, performing evaluations at the specified boundaries. The exercise showing the area under \(x^2\) uses this theorem to transition from finding an antiderivative to evaluating the integral with ease.