Problem 19 - 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. \(f(x)=12 x^{2}+5 \quad a=0, b=3\)

Step-by-Step Solution

Answer

The average value is 41, and \\(c = \sqrt{3} \approx 1.732\\). Geometrically, this is the rectangle height matching the curve's area under \\(f(x)\\) from 0 to 3.

Step 1: Calculate the Integral of the Function

To determine the average value of the function over the interval \([a, b] = [0, 3]\), start by computing the integral of \(f(x) = 12x^2 + 5\) over the interval. Use the formula: \[\int_{a}^{b} f(x)\, dx = \int_{0}^{3} (12x^2 + 5)\, dx\]Integrate each term separately:\[\int 12x^2\, dx + \int 5\, dx = 12\left(\frac{x^3}{3}\right) + 5x \bigg|_0^3\]Calculate using the boundaries:\[= 12\left(\frac{3^3}{3}\right) + 5\times3 - \left(12\frac{0^3}{3} + 5\times0\right) = 108 + 15 = 123\]

Step 2: Determine the Average Value of the Function

The average value \(f_{\text{avg}}\) of \(f(x)\) over the interval \([0, 3]\) is given by the formula:\[f_{\text{avg.}} = \frac{1}{b-a} \int_{a}^{b} f(x)\, dx = \frac{1}{3-0} \times 123 = \frac{123}{3} = 41\]So, the average value of the function \(f(x)\) over the interval \([0, 3]\) is 41.

Step 3: Find the Point c Where f(c) = f_{avg}

We need to find a point \(c\) in \((0, 3)\) where \(f(c) = f_{\text{avg}} = 41\). Set the function equal to 41:\[12c^2 + 5 = 41\]Solve for \(c\):\[12c^2 = 41 - 5\]\[12c^2 = 36\]\[c^2 = 3\]\[c = \sqrt{3} \approx 1.732\]Therefore, the point \(c\) where \(f(c) = f_{\text{avg}}\) is \(c = \sqrt{3}\approx 1.732\).

Step 4: Illustrate the Geometric Significance of c

The point \(c\) where \(f(c) = f_{\text{avg.}}\) represents the height of a rectangle that has the same area as the area under the curve from \(a = 0\) to \(b = 3\). Geometrically, if you were to draw the function and a rectangle that spans from \(x = 0\) to \(x = 3\), the rectangle's height would be 41 and would touch the curve at \(x = \sqrt{3}\). This demonstrates that the entire area under the curve is equivalent to the area of this rectangle.

Key Concepts

Calculating Integrals

To find the average value of a function over a given interval, the first step is to compute the definite integral of the function over that interval. Integrals allow us to calculate the area under a curve, which is essential to understanding how the function behaves over a specific range.
In this problem, our task was to integrate the function \( f(x) = 12x^2 + 5 \) from \( a = 0 \) to \( b = 3 \). We apply the following integral formula:

• \( \int_{a}^{b} f(x)\, dx = \int_{0}^{3} (12x^2 + 5)\, dx \)

We calculate this by integrating each term separately. For the \( 12x^2 \) term, we use the power rule to get \( 12 \times \frac{x^3}{3} \). For the constant 5, integrating straightforwardly yields \( 5x \). Evaluating these from 0 to 3 gives us a total area of 123 under the curve. This area plays a crucial role in determining the function's average value.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the processes of differentiation and integration, providing a powerful tool for analysis. The theorem states that if a function is continuous over an interval \([a, b]\), then integrating that function over \([a, b]\) and differentiating it will return to the original function.
In simpler terms, it tells us that the derivative of an integral of a function is the function itself. In this exercise, once we've found the integral of the function over the interval, we divide this integral by the interval's length to find the average value:

• \( f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x)\, dx \)
• Here, \( f_{\text{avg}} = \frac{123}{3} = 41 \)

This gives a clear understanding of the function's average behavior over the interval. Observing this average value provides insight into the general level at which the function operates over the given range.

Geometric Interpretation

To visualize the meaning of the average value, imagine the function drawn as a curve from \(x = 0\) to \(x = 3\). The average value then corresponds to a horizontal line that cuts across the height of 41. This line indicates the level at which the area under the curve would be uniformly stretched across the entire interval.
Finding the specific point \( c \, (\sqrt{3} \approx 1.732) \) at which the function value exactly equals this average reflects where this height visually meets the curve. At this point, drawing a rectangle with a base from 0 to 3 and a height of 41 would match the area under the curve.
This geometric reflection helps cement the concept that the average value measures 'evenness' of distribution, making the notion of averaging accessible and visually intuitive.