Problem 3 - Active Calculus

Question

Find the average value of \(f(x)=6 x+5\) over [2,6] average value = _____________

Step-by-Step Solution

Answer

The average value is 29.

Step 1: Identify the formula for the average value

The average value of a function over the interval \([a, b]\) is given by \(\frac{1}{b-a} \int_a^b f(x) \, dx\). For this problem, \[a = 2\text{ and }b = 6\text{ and }f(x) = 6x + 5.\]

Step 2: Set up the integral

Set up the definite integral \(\frac{1}{6-2} \int_2^6 (6x + 5) \, dx\). This simplifies to \(\frac{1}{4} \int_2^6 (6x + 5) \, dx\).

Step 3: Find the antiderivative

Compute the antiderivative of \(6x + 5\). The antiderivative is \(3x^2 + 5x\).

Step 4: Evaluate the definite integral

Evaluate the antiderivative from \2 \text{ to } 6\: \[: 3x^2 + 5x\]_2^6\. Substituting in the bounds, we get\[ (3(6^2) + 5(6)) - (3(2^2) + 5(2))\].

Step 5: Calculate the result

Perform the arithmetic: \( (3(36) + 30) - (3(4) + 10) = (108 + 30) - (12 + 10) = 138 - 22 = 116\).

Step 6: Divide by the interval length

Divide by the length of the interval \([a, b]\): \( \frac{116}{4} = 29\).

Key Concepts

definite integral

The definite integral is a fundamental concept in calculus that represents the area under a curve between two points on the x-axis. For a given function, the definite integral from point \( a \) to point \( b \) is denoted as \( \int_a^b f(x) \, dx \). This integral is computed by evaluating the area under the curve of \( f(x) \) between the limits \( a \) and \( b \). It's important to understand that the definite integral provides a numerical value representing this area. For our exercise, the definite integral we set up was \( \int_2^6 (6x + 5) \, dx \), which helps us in finding the average value of the given function over the specified interval.

antiderivative

An antiderivative, also known as an indefinite integral, is the reverse process of differentiation. It helps in finding the original function from its derivative. In the context of definite integrals, the antiderivative is used to find the area under the curve. For instance, to find the average value of the function \( f(x) = 6x + 5 \), we first need its antiderivative. For \(6x + 5\), the antiderivative is \( 3x^2 + 5x \). This is obtained by reversing the differentiation process, where we integrate each term of the given function separately.

interval length division

To find the average value of a function over an interval, we use the definite integral and then divide by the length of the interval. This step ensures we normalize the total area under the curve by the width of the interval, providing the mean value over that span. Mathematically, this is expressed as: \( \text{{Average value}} = \frac{1}{b-a} \int_a^b f(x) \, dx \). For our exercise:

• The interval is \([2, 6]\), thus \( b - a = 6 - 2 = 4 \).
• After evaluating the integral, we divide the resulting area by 4. This calculation yields the average value of the function \( 6x + 5 \) over the interval \([2, 6]\).

evaluation of definite integrals

Once we have computed the antiderivative of our function, the next step is to evaluate it within the bounds of the interval. This involves substituting the upper and lower limits into the antiderivative and subtracting the latter from the former. For our function \( 6x + 5 \):

• The antiderivative is \( 3x^2 + 5x \).
• We evaluate it from 2 to 6, substituting these values gives us \( (3(6^2) + 5(6)) - (3(2^2) + 5(2)) \).
• This computation simplifies to \( (108 + 30) - (12 + 10) \), resulting in \( 138 - 22 = 116 \). Finally, we divide by the interval length (4) to find the average value: \( \frac{116}{4} = 29 \).