Question

(a) Find \(f_{\text {avg }}\) of \(f(x)=x^{2}\) over \([0,2]\). (b) Find a number c in \([0,2]\) such that \(\mathrm{f}(\mathrm{c})=\mathrm{f}_{\text {avg' }}\) (c) Sketch the graph of \(f(x)=x^{2}\) over \([0,2]\) and construct a rectangle over the interval whose area is the same as the area under the graph of fover the interval.

Short answer

Based on the given step by step solution, the average value of the function \(f(x)=x^2\) over the interval \([0, 2]\) is \(\frac{4}{3}\). The number \(c\) for which \(f(c)=f_{\text{avg}}\) is \(c=\sqrt{\frac{4}{3}}\).

Step-by-step solution

Calculate average value of \(f(x)\)

To find the average value of the function \(f(x) = x^2\) over the interval \([0, 2]\), we can use the formula: $$f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx$$ where \(a\) and \(b\) are the endpoints of the interval \([0, 2]\). Here, \(a = 0\), \(b = 2\), and \(f(x) = x^2\). Plugging these values into the formula, we get: $$f_{\text{avg}} = \frac{1}{2-0} \int_0^2 x^2 \, dx$$ To find the integral, we apply the power rule and obtain: $$\int_0^2 x^2 \, dx = \frac{1}{3}x^3 \biggr|_0^2$$ Calculating the value at the boundaries, we obtain: $$\frac{1}{3}(2^3) - \frac{1}{3}(0^3) = \frac{1}{3}(8) = \frac{8}{3}$$ Now, substitute this value back into the \(f_{\text{avg}}\) formula to get: $$f_{\text{avg}} = \frac{1}{2}(\frac{8}{3}) = \frac{4}{3}$$

Find the value of c

We need to find a number \(c\) in the interval \([0, 2]\) such that \(f(c) = f_{\text{avg}}\). We already know that \(f_{\text{avg}} = \frac{4}{3}\), so we have the equation: $$f(c) = \frac{4}{3}$$ Now, we must solve for \(c\): $$c^2 = \frac{4}{3}$$ $$c = \sqrt{\frac{4}{3}}$$ Since \(c\) is in the interval \([0, 2]\), we found the desired value.

Sketch the graph and construct the rectangle

Sketch the graph of \(f(x) = x^2\) over the interval \([0, 2]\). To construct the rectangle with the same area as the area under the graph of \(f(x)\), we will use the coordinates \((0, 0)\), \((2, 0)\), \((0, f_{\text{avg}})\), and \((2, f_{\text{avg}})\) which are: \((0, 0)\), \((2, 0)\), \((0, \frac{4}{3})\), and \((2, \frac{4}{3})\) The height of the rectangle is equal to \(f_{\text{avg}}\), which is \(\frac{4}{3}\). So, the area of the rectangle will be equal to \(2 \times \frac{4}{3} = \frac{8}{3}\), which is the area under the graph of \(f(x)=x^2\) over the interval \([0, 2]\). After sketching the graph and the rectangle, the completed graph should visually represent the average value of the function and the area under the curve for the given interval.

Key concepts

Definite Integration

Definite integration plays a fundamental role in calculating the area under a curve for a given function over a specified interval. At its core, definite integration takes a function, like our example function \( f(x) = x^2 \), and determines the total accumulation of the function's values from one point to another.

Returning to the original exercise, we see definite integration in action. The process starts by finding the integral of \( x^2 \) from 0 to 2. This is symbolically written as \( \int_0^2 x^2 \, dx \). The notation \( dx \) signifies that the function \( x^2 \) is being integrated with respect to \( x \) across the interval from 0 to 2. After applying the power rule of integration, we obtain \( \frac{1}{3}x^3 \biggr|_0^2 \), which computes the antiderivative at the upper and lower limits.

The result of this operation gives us the exact area under the curve of the function from 0 to 2, which is an integral aspect of understanding how function behavior is quantified and visualized.

Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals provides the insight that for a continuous function over a closed interval, there exists at least one point at which the function value equals the average value of the function over that interval. This theorem links our abstract concept of an average to a concrete value that the function actually attains.

In the solution for the original exercise, we calculated the average value of the function \( f(x) = x^2 \) using the formula \( f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \). After determining the average value to be \( \frac{4}{3} \), we then sought a specific point \( c \) within our interval \( [0,2] \), which satisfies the equation \( f(c) = f_{\text{avg}} \). Solving this leads to discovering the number \( c = \sqrt{\frac{4}{3}} \). Here, the mean value theorem is beautifully exemplified, showing that such a point \( c \) does indeed exist, thus connecting the abstract average value with a tangible point on the graph of the function.

Graphical Representation of Functions

The graphical representation of functions is a visual interpretation of the relationship between a function's input values and its outputs. It's a tool that significantly aids in understanding functions' behavior. The importance of graphing, as shown in the third part of the exercise, is in turning numeric data and operations into visual information, which helps to deepen understanding and intuition.

In our case, the exercise required sketching the graph of \( f(x) = x^2 \) over the interval \( [0, 2] \). By creating a smooth parabolic curve from \((0, 0)\) to \((2, 4)\), we achieve a visual representation of the function. To reinforce the concept of average value, we were instructed to construct a rectangle within the same interval with a height of \( f_{\text{avg}} = \frac{4}{3} \), which gives the rectangle the same area as the region under the curve. The resulting image not only aids in comprehension but also provides a graphical proof of the mean value theorem for integrals, showing its value on the actual function graph.