Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

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 \( c \) that satisfies the Mean Value Theorem for Integrals is \( c = 2.5 \).

Step 1: Mean Value Theorem for Integrals Formula

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

Step 2: Calculate the Integral

First, calculate the definite integral of \( f(x) = ax + b \) from 1 to 4. Using the power rule, the integral of \( ax \) is \( \frac{ax^2}{2} \) and the integral of \( b \) is \( bx \). The definite integral is: \[ \int_1^4 (ax + b) \,dx = \left[ \frac{ax^2}{2} + bx \right]_1^4. \]

Step 3: Evaluate the Integral

Evaluate the integral using the Fundamental Theorem of Calculus:\[ \left( \frac{a(4)^2}{2} + b(4) \right) - \left( \frac{a(1)^2}{2} + b(1) \right). \]Calculating these terms gives: \[ \left( 8a + 4b \right) - \left( \frac{a}{2} + b \right) = \frac{15a}{2} + 3b. \]

Step 4: Substitute into the Mean Value Theorem

Substitute the evaluated integral back into the Mean Value Theorem formula:\[ f(c) = \frac{1}{3} \left( \frac{15a}{2} + 3b \right) = 5a/2 + b. \]

Step 5: Solve for c

Since \( f(x) = ax + b \), then \( f(c) = ac + b \). Equate this to the result from Step 4 and solve for \( c \):\[ ac + b = \frac{5a}{2} + b. \]Subtract \( b \) from both sides:\[ ac = \frac{5a}{2}. \]Divide both sides by \( a \) (assuming \( a eq 0 \)):\[ c = \frac{5}{2}. \]

Step 6: Verify c is in the Interval

Verify that \( c = \frac{5}{2} = 2.5 \) lies within the interval \([1,4]\). Since it does, \( c = 2.5 \) is a valid solution.

Key Concepts

Continuous Function

A continuous function is a type of function that is unbroken and smooth, meaning that it can be drawn on a paper without lifting the pencil. When working with the Mean Value Theorem for Integrals, ensuring that the function is continuous over the given interval is crucial.
This is because discontinuities can create jumps or gaps in the graph, causing issues when calculating integrals.
Continuous functions exhibit the following properties:

• For any small change in the input, there is a corresponding small change in the output.
• There are no sudden jumps, breaks, or holes in the graph of the function.

In simple terms, a continuous function has no unexpected surprises. For the function given in the exercise, such as a linear function like \(f(x) = ax + b\), it is always continuous over any interval, including the interval \([1, 4]\). This scope of continuousness assures that we can safely apply the integral and the Mean Value Theorem.

Definite Integral

The concept of a definite integral can be thought of as the net area under a curve, from one point to another on the x-axis.
For linear functions, such as \(f(x) = ax + b\), it involves calculating the total accumulation of values from starting point \(a\) to endpoint \(b\) on the given interval.
Here's how it fits into the Mean Value Theorem:

• We calculate it using the notation \(\int_a^b f(x) \, dx\). The result is a numerical value representing the total change.
• In the original solution, we used the definite integral from 1 to 4.
• This calculation involved using the power rule to find the antiderivative and then evaluating it at the upper and lower limits.

In our exercise, the definite integral gave us \(\frac{15a}{2} + 3b\), a result that became a pivotal part of finding \(c\) in the theorem. Integrals are therefore essential when understanding how values accumulate over an interval.

Fundamental Theorem of Calculus

One of the cornerstone ideas in calculus, the Fundamental Theorem of Calculus, connects derivatives and integrals.
It gives us a powerful way to evaluate definite integrals and it is divided into two parts.

• The first part assures the existence of antiderivatives for continuous functions, meaning integrals can be connected back to an actual function.
• The second part, more relevant for definite integrals, allows us to find the accumulated value from point \(a\) to \(b\) by subtracting the value of the antiderivative at \(a\) from that at \(b\).

For the linear function \(ax + b\), the Fundamental Theorem of Calculus simplified our task by enabling us to evaluate the integral directly using the antiderivative \(\frac{ax^2}{2} + bx\).
This provided a straightforward approach to validate our result of \(\frac{15a}{2} + 3b\), crucial for applying the Mean Value Theorem for Integrals.