Question

Mean Value Theorem for Integrals Find or approximate all points at which the given function equals its average value an the given interval. $$f(x)=8-2 x \text { on }[0,4]$$

Short answer

Question: On the interval [0, 4], find the point(s) at which the function f(x) = 8 - 2x equals its average value. Answer: The function f(x) equals its average value at x = 2.

Step-by-step solution

Determine the average value of the function

We will first calculate the integral of the given function over the given interval [0, 4] and then use this value to calculate the average value of the function. To find the integral of f(x) = 8 - 2x, we first compute the antiderivative F(x) of f(x): $$F(x) = \int f(x) dx = 8x - x^2 + C$$ However, since we are computing a definite integral, the constant C would cancel out when evaluating F(b) - F(a). Therefore, we do not need to worry about adding a constant. Now we calculate the integral of f(x) over the interval [0, 4]: $$\int_{0}^{4} f(x) dx=\Big[8x - x^2\Big]_0^4 = (8(4) - 4^2) - (8(0) - 0^2) = 16$$ Next, we will calculate the average value of the function by dividing the integral by the length of the interval (4-0), which is 4: $$\text{average value} = \frac{1}{4-0} \int_{0}^{4} f(x) dx = \frac{1}{4} \cdot 16 = 4$$

Solve the equation f(c) = average value

Now we need to find all points c for which f(c) equals the average value of 4. To do this, we set up the following equation and solve for c: $$f(c) = 4 \Rightarrow 8 - 2c = 4$$ Adding 2c to both sides and subtracting 4 from both sides yields: $$2c = 8 - 4 \Rightarrow 2c = 4$$ Finally, we divide by 2 to get the value of c: $$c = \frac{4}{2} = 2$$ The only point at which the function f(x) equals its average value on the interval [0, 4] is at \(x = 2\).

Key concepts

Definite Integral Calculation

When we talk about definite integral calculation, we refer to the process of determining the exact area under the curve of a function between two specific points on the x-axis. This concept is central in calculus and provides insights into various physical and mathematical phenomena such as displacement, area, and volume.

In the context of our exercise, to calculate the definite integral of the function f(x) = 8 - 2x from x = 0 to x = 4, we evaluate the integral at the upper limit and subtract the value of the integral at the lower limit. This is compactly written as \[\int_{0}^{4} f(x) dx = [F(x)]_0^4\] where F(x) is the antiderivative of f(x). The calculation yields 16, which represents the area under the curve f(x) within the interval [0, 4].

Average Value of a Function

The average value of a function over an interval offers a single representative number that summarizes the behavior of the function over that range. To find it, you divide the area under the curve by the width of the interval.

In our exercise, after calculating the definite integral, we find the average value by dividing this result (16) by the length of the interval (4). This computation, \[\text{average value} = \frac{1}{b-a} \int_{a}^{b} f(x) dx\], where a and b are the interval limits, gives us 4. It means that if the function f(x) could be evenly distributed along the y-axis, it would consistently hit the mark of 4 across the entire interval.

Antiderivative

An antiderivative, often called the indefinite integral, is the reverse process of differentiation. When we find an antiderivative for a function, we are seeking a function whose derivative is the original function. It's important to note that antiderivatives are not unique because any constant added to an antiderivative is also an antiderivative; this constant is represented by C in expressions.

In our exercise, the function F(x) = 8x - x^2 is an antiderivative of f(x) = 8 - 2x. Whenever we compute a definite integral, we utilize an antiderivative to simplify the calculation. However, for definite integrals, the constant C becomes irrelevant since it cancels out when subtracting the lower bound from the upper bound.

Function Analysis

Function analysis involves examining properties of a function, such as its limits, continuity, differentiability, and behavior over intervals of interest. In the practical application of our exercise, function analysis includes identifying where the function reaches its average value.

We set the function f(x) equal to its average value and solve for x. By analyzing the function and using algebraic manipulation, we discover that f(x) equals its average value at x = 2. This intersection point gives us valuable information about the function's behavior on the interval [0, 4].