Problem 16 - Calculus Single Variable

Question

Compute the average value \(f_{\text {avg }}\) of \(f\) over \([a, b]\) and find a point \(c\) in \((a, b)\) for which \(f(c)=f_{\text {avg. }}\) Illustrate the geometric significance of \(c\) with a sketch accompanied by a description. \(f(x)=x^{3} \quad a=0, b=2\)

Step-by-Step Solution

Answer

The average value is 2, and the point \( c = \sqrt[3]{2} \approx 1.26 \) satisfies \( f(c) = f_{\text{avg}} \).

Step 1: Define the Average Value Formula

The average value of a function \( f \) over the interval \([a, b]\) is given by \[ f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] We will use this formula to find the average value of the function \( f(x) = x^3 \) over the interval \([0, 2]\).

Step 2: Compute the Integral

Evaluate the integral \( \int_0^2 x^3 \, dx \) to find the total area under the curve from \( a = 0 \) to \( b = 2 \). The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \), so \[ \int_0^2 x^3 \, dx = \left. \frac{x^4}{4} \right|_0^2 = \frac{(2)^4}{4} - \frac{(0)^4}{4} = \frac{16}{4} = 4. \]

Step 3: Calculate the Average Value

Substitute the result of the integral into the average value formula: \[ f_{\text{avg}} = \frac{1}{2-0} \times 4 = \frac{4}{2} = 2. \] The average value of the function \( f(x) = x^3 \) over the interval \([0, 2]\) is \( 2 \).

Step 4: Find the Point \( c \)

We need to find a value \( c \) in the interval \((0, 2)\) such that \( f(c) = f_{\text{avg}} = 2 \). Solve \( c^3 = 2 \), i.e., \[ c = \sqrt[3]{2}. \] Thus, \( c \approx 1.26 \).

Step 5: Illustrate the Geometric Significance

The average value \( f_{\text{avg}} = 2 \) represents a horizontal line across the graph of \( f(x) = x^3 \) from \( x = 0 \) to \( x = 2 \). The point \( c = \sqrt[3]{2} \) is where the curve intersects with this horizontal line, indicating where the function's instantaneous output is equal to its average output over the interval. Thus, \( c \) geometrically balances the area under \( f(x) \).

Step 6: Sketch for Understanding

On a graph, plot the curve \( y = x^3 \) over the interval \([0, 2]\), and draw the horizontal line \( y = 2 \). Mark the point \( c = \sqrt[3]{2} \) on this line where it intersects the curve. This visualization shows that the area below and above the line up to this point \( c \) is equal, emphasizing the balance.

Key Concepts

Integral Calculus

Integral calculus plays a crucial role in understanding how a function behaves over a range. More specifically, it helps in determining the total accumulation of quantities, like areas under curves. When we talk about computing the average value of a function, we are essentially using integration to sum up all the infinitesimal values of the function over an interval. This cumulative essence is captured by the integral, which computes the area under the curve of the function from point \(a\) to point \(b\).

• The integral of \(f(x) = x^3\) from 0 to 2 is calculated as \(\int_0^2 x^3 \, dx\).
• This integral equals 4, representing the total area under the curve over the interval \([-\infty, \infty]\).

By dividing this integral by the length of the interval, as per the average value formula, we can find the mean output of the function over this span.

Antiderivative

The concept of an antiderivative refers to the function whose derivative is the given function. When we compute an integral, we effectively find the antiderivative of the function involved. For \(f(x) = x^3\), the antiderivative is \(\frac{x^4}{4}\). This means:

• If you differentiate \(\frac{x^4}{4}\), you return to \(x^3\).
• We use the Fundamental Theorem of Calculus to evaluate the definite integral by plugging in the bounds of integration, in this case, from 0 to 2.

This process allows us to find the total area under the curve of \(x^3\) from 0 to 2 by evaluating \(\frac{x^4}{4} \bigg|_0^2\), which results in 4. Antidifferentiation is essential for solving integrals and understanding cumulative change.

Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals bridges the concepts of average values and continuous functions over an interval. This theorem states that if \(f(x)\) is continuous over \([a, b]\), there exists a point \(c\) in \((a, b)\) where the function's instantaneous value is equal to its average value.

• For the function \(f(x) = x^3\), its average value over \([0, 2]\) is 2.
• To find the point \(c\), we solve \(c^3 = 2\), giving \(c = \sqrt[3]{2} \approx 1.26\).

This point \(c\) confirms a place between 0 and 2 where the output of \(x^3\) equals its mean value over the interval, illustrating the hallmark of this theorem.

Geometric Interpretation

Visualizing mathematical concepts often provides clearer understanding. In this situation, the average value found is 2, represented graphically by a horizontal line at \(y = 2\) extending from \(x = 0\) to \(x = 2\).

• The curve \(y = x^3\) rises from \(x = 0\) to \(x = 2\), crossing this horizontal line once.
• The intersection point, \(c = \sqrt[3]{2} \approx 1.26\), marks where the function's height equals the average value.

This intersection represents a balance point, or equilibrium, where the area above and below this line are equal over the entire interval. Sketching both the function and the average line allows one to appreciate how the area under the curve up to \(c\) equals the remaining area over the interval up to 2, solidifying the concept of function means geometrically.