Problem 23 - 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. $$ R(v)=v^{2}-v ; \quad[0,2] $$

Step-by-Step Solution

Answer

The value of \( c \) that satisfies the theorem on \([0,2]\) is \( c = \frac{3 + \sqrt{33}}{6} \approx 1.45 \).

Step 1: Mean Value Theorem for Integrals Formula

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 number \( c \) in \((a, b)\) such that:\[ f(c) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]

Step 2: Calculate the Integral

Evaluate the integral of \( R(v) = v^2 - v \) from \(0\) to \(2\):\[ \int_{0}^{2} (v^2 - v) \, dv = \left[ \frac{v^3}{3} - \frac{v^2}{2} \right]_{0}^{2}\]This simplifies to:\[ \left( \frac{2^3}{3} - \frac{2^2}{2} \right) - \left( \frac{0^3}{3} - \frac{0^2}{2} \right) = \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}\]

Step 3: Set Up Equation for f(c)

The Mean Value Theorem states:\[ R(c) = \frac{1}{2-0} \int_{0}^{2} R(v) \, dv = \frac{2}{3}\]So, we need:\[ c^2 - c = \frac{2}{3}\]

Step 4: Solve for c

To solve \( c^2 - c = \frac{2}{3} \), first clear the fraction by multiplying the entire equation by 3:\[ 3c^2 - 3c = 2\]Rearrange to a standard quadratic form:\[ 3c^2 - 3c - 2 = 0\]Apply the quadratic formula:\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a = 3, b = -3, c = -2 \), thus:\[ c = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} = \frac{3 \pm \sqrt{9 + 24}}{6} = \frac{3 \pm \sqrt{33}}{6}\]

Step 5: Evaluate Solutions

Calculate \( c \) values:\[ c_1 = \frac{3 + \sqrt{33}}{6}, \quad c_2 = \frac{3 - \sqrt{33}}{6}\]Since \( c \) must lie within the interval \((0,2)\), evaluate each solution:Given that \( \sqrt{33} \approx 5.74 \):- \( c_1 = \frac{3 + 5.74}{6} \approx 1.45 \) which is in the interval.- \( c_2 = \frac{3 - 5.74}{6} \approx -0.457 \) is not in the interval.Therefore, only \( c_1 \) is valid.

Key Concepts

Continuous Function

In mathematics, a function is called continuous if there are no abrupt changes in its value. This means the graph of the function can be drawn without lifting the pen from the paper. Continuous functions play a crucial role in many areas of calculus, including the Mean Value Theorem for Integrals.

The Mean Value Theorem for Integrals applies exclusively to continuous functions on a given interval

• This theorem guarantees that there exists at least one point in the interval where the average value of the function equals the function's value at that point.
• In the original problem, the function given is a polynomial: \( R(v)=v^{2}-v \). Polynomials are always continuous on any interval of real numbers.

With this assurance of continuity, we can apply integral calculus to find particular values that satisfy certain conditions, as seen in the original exercise.

Integral Calculus

Integral calculus is a branch of calculus focusing on the concept of integration, which is essentially about finding the whole given a rate of change. In the context of the Mean Value Theorem for Integrals, it helps in determining the average value of a function over a specified interval.

To apply this theorem, you need to calculate the definite integral of the given function over the specified interval. The integration process can be likened to summing up an infinite number of infinitely small parts to find the total. For the function in question, \( R(v) = v^2 - v \), the integral from 0 to 2 is calculated as:

• Firstly, find the antiderivative, which involves determining a function whose derivative is the given function.
• Then, evaluate this antiderivative at the bounds of the interval (from 0 to 2 in this case).
• Subsequently, subtract the value of the antiderivative at the lower bound from its value at the upper bound.

This calculates the area under the curve of the function, crucial for computing average values using the Mean Value Theorem for Integrals.

Quadratic Equation Solving

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). To solve these, we use the quadratic formula: \[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula provides solutions in terms of the coefficients of the equation, where \( a \), \( b \), and \( c \) are constants.

In our example stemming from the application of the Mean Value Theorem for Integrals, we end up with a quadratic equation in \( c \):

• The function \( R(c) = \frac{2}{3} \) transforms into \( c^2 - c = \frac{2}{3} \).
• By multiplying through by 3 to clear the fraction, this becomes \( 3c^2 - 3c - 2 = 0 \).
• Using the quadratic formula, we solve for \( c \), calculating possible solution values.

The solutions must then be checked to ensure they lie within the given interval. Only solutions which are valid in the context of the initial problem (i.e., within the specific range given) are considered correct.