Problem 25 - Calculus

Question

Find all values of c that satisfy the Mean Value Theorem for Integrals on the given interval. $$ f(x)=a x+b ; \quad[1,4] $$

Step-by-Step Solution

Answer

The value of \( c \) satisfying the theorem is \( c = \frac{5}{2} \).

Step 1: Understand the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \( f(x) \) is continuous on the interval \([a, b]\), then there exists at least one point \( c \) in the interval \((a, b)\) such that \( f(c) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \).

Step 2: Compute the Integral of \( f(x) \) from 1 to 4

We need to calculate \( \int_{1}^{4} (ax + b) \, dx \). The antiderivative of \( ax + b \) is \( \frac{a}{2}x^2 + bx \). Evaluating this from 1 to 4, we have: \[ \left. \frac{a}{2}x^2 + bx \right|_{1}^{4} = \left( \frac{a}{2}(4)^2 + b(4) \right) - \left( \frac{a}{2}(1)^2 + b(1) \right) \] \[ = \left( 8a + 4b \right) - \left( \frac{a}{2} + b \right) \].

Step 3: Simplify the Integral Result

Simplify the expression: \[ (8a + 4b) - \left( \frac{a}{2} + b \right) = 8a + 4b - \frac{a}{2} - b \] \[ = \frac{16a}{2} + \frac{8b}{2} - \frac{a}{2} - \frac{2b}{2} \] \[ = \frac{15a}{2} + \frac{6b}{2} = \frac{15a + 6b}{2} \].

Step 4: Apply the Mean Value Theorem

Using the theorem, set \( f(c) = \frac{1}{b-a} \int_{1}^{4} f(x) \, dx \):\( a(c) + b = \frac{1}{4-1} \times \frac{15a + 6b}{2} = \frac{15a + 6b}{6} \).Multiply through by 6 to clear the fraction:\[ 6(ac + b) = 15a + 6b \].

Step 5: Solve the Equation for c

Distribute and rearrange the equation:\( 6ac + 6b = 15a + 6b \).Cancel out \( 6b \) on both sides:\( 6ac = 15a \).Divide by \( 6a \) (assuming \( a eq 0 \)):\( c = \frac{15}{6} = \frac{5}{2} \).

Step 6: Finalize the Solution

The value \( c = \frac{5}{2} \) lies within the open interval \((1, 4)\), thus satisfying the conditions of the Mean Value Theorem for Integrals.

Key Concepts

Understanding Continuous Functions

A continuous function is one of the essential requirements for the Mean Value Theorem for Integrals. A function is said to be continuous if, intuitively, you can draw it without lifting your pen off the paper. Mathematically, it means there are no breaks, jumps, or holes in the graph of the function on the specified interval. For our problem, the function \(f(x) = ax + b\) is a linear function. Linear functions are inherently continuous everywhere on the real number line. Thus, the required condition of continuity on the interval \([1, 4]\) is naturally satisfied.When dealing with such problems, ensuring the function is continuous on the given interval is critical, as it guarantees that the Mean Value Theorem can be applied.

Integral Calculation Essentials

Integral calculation involves finding the area under the curve of a graph represented by a function. For definite integrals, you're interested in the signed area between the curve and the x-axis over a specific interval. In the exercise, we computed \(\int_{1}^{4} (ax + b) \, dx\). This involved finding the antiderivative first: \(\frac{a}{2}x^2 + bx\). Once the antiderivative is determined, you evaluate it at the inteval bounds: first at 4, then at 1. The calculation yields \(\frac{15a + 6b}{2}\), representing the area under the curve between \(x=1\) and \(x=4\). Note that this integral expression is crucial for applying the theorem later.

Exploring the Antiderivative

The antiderivative, often called the indefinite integral, is a fundamental concept needed for solving definite integrals. Essentially, it's the reverse process of differentiation, where you find a function whose derivative matches the original function.For the function \(f(x) = ax + b\), the antiderivative is calculated to be \(\frac{a}{2}x^2 + bx\). This represents all functions \(F(x)\) such that when you differentiate \(F(x)\), you get \(f(x)\). Finding the antiderivative is a key step in integral calculation, allowing you to employ the Fundamental Theorem of Calculus, which connects differentiation and integration, to evaluate definite integrals effectively.

Defining the Interval of Interest

An interval is a range of values within which you analyze functions for properties like continuity, or calculate integrals. Here, the given interval \([1, 4]\) is vital for not only calculating the integral but also for checking conditions of the Mean Value Theorem for Integrals.The interval \([1, 4]\) indicates where to assess the function's behavior. It defines the limits for integration, and specifies where the required value \(c\) must lie. The theorem assures that such a \(c\) exists within the open interval \((1, 4)\), if the function is continuous there. Understanding intervals helps in focusing analysis on specific regions and ensuring all conditions are met for theorems to apply correctly.