Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 15-28, find all values of \(c\) that satisfy the Mean Value Theorem for Integrals on the given interval. 15\. \(f(x)=\sqrt{x+1} ; \quad[0,3] \quad\)

Step-by-Step Solution

Answer

The value of \(c\) is \(\frac{115}{81}\).

Step 1: Recall the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \(f\) is continuous on the closed interval \([a, b]\), there exists at least one point \(c\) in \( (a, b) \) such that:\[f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx\]First, ensure that the given function \(f(x) = \sqrt{x+1}\) is continuous over the interval \([0, 3]\). Since \(f(x)\) is a square root function shifted horizontally, it is continuous everywhere its argument is non-negative, which it is on \([0, 3]\).

Step 2: Compute the Definite Integral

Calculate the integral of \(f(x) = \sqrt{x+1}\) from 0 to 3:\[\int_0^3 \sqrt{x+1} \, dx\]To solve this, use the substitution method. Let \(u = x + 1\), then \(du = dx\) and the limits of integration become from 1 to 4:\[\int_1^4 \sqrt{u} \, du = \left[ \frac{2}{3}u^{3/2} \right]_1^4\]Evaluating this gives:\[\frac{2}{3} \times (4^{3/2} - 1^{3/2}) = \frac{2}{3} \times (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3}\]

Step 3: Calculate the Average Value of the Function

Apply the Mean Value Theorem by calculating the average value of the function over the interval. This is given by:\[\frac{1}{3-0} \int_0^3 \sqrt{x+1} \, dx = \frac{1}{3} \times \frac{14}{3} = \frac{14}{9}\]

Step 4: Solve for \(c\)

We need \(f(c) = \frac{14}{9}\). Therefore:\[\sqrt{c+1} = \frac{14}{9}\]Square both sides to remove the square root:\[c+1 = \left(\frac{14}{9}\right)^2 = \frac{196}{81}\]Subtract 1 from both sides:\[c = \frac{196}{81} - 1 = \frac{196}{81} - \frac{81}{81} = \frac{115}{81}\]

Step 5: Verify \(c\) is within the Interval

Check if the calculated \(c = \frac{115}{81}\) is within the open interval \((0, 3)\). Convert to decimal to verify:\[\frac{115}{81} \approx 1.42\]Since \(1.42\) is within \(0 < c < 3\), the value is valid.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the accumulated area under a curve from one point to another on a specific interval. The notation for a definite integral is \( \int_a^b f(x) \, dx \), where \(a\) and \(b\) are the lower and upper bounds, respectively. To calculate the definite integral of a function, you determine the total sum of the infinitely small areas under the curve of the function between these bounds.

• The process involves evaluating an integral over a specific interval \([a, b]\).
• Definite integrals provide exact quantities, unlike indefinite integrals that give general antiderivatives.

In the given problem, we calculated the definite integral of \(\sqrt{x+1}\) over the interval \([0,3]\) using the substitution method. This integral resulted in the value \(\frac{14}{3}\), which was used to find the average value of the function on this interval.

Continuous Functions

Continuous functions are the ones that have no breaks, jumps, or holes in their graphs. For a function to be continuous on an interval, it must be continuous at every point within that interval. In the context of the Mean Value Theorem for Integrals, continuity is crucial as it guarantees the existence of at least one value \(c\) that satisfies the equality given by the theorem.

• Continuous functions have predictable behavior and are essential in calculus to derive meaningful results from integrals.
• The function \(f(x) = \sqrt{x+1}\) is continuous on \([0,3]\) because the square root function is defined and smooth, as long as its argument is non-negative.

This means that within the interval \([0,3]\), the function behaves nicely without any disruptions, making it possible to apply the Mean Value Theorem for Integrals successfully.

Substitution Method

The substitution method is a technique used in calculus to simplify evaluating integrals. By substituting one part of the integrand with a new variable, you can transform the integral into a simpler form, which is often easier to solve. It is similar to the chain rule for derivatives but applied in reverse.

• To use this method, select a new variable \(u\) to replace a complex part of the integrand.
• Don't forget to change the limits of integration according to the new variable.

For the function \(f(x) = \sqrt{x+1}\), we use substitution here by letting \(u = x+1\), leading to \(du = dx\). This changes the limits of the integral and transforms \(\int_0^3 \sqrt{x+1} \, dx\) into \(\int_1^4 \sqrt{u} \, du\), simplifying the process of finding the value of the definite integral.

Average Value of a Function

The average value of a function on an interval gives a single representative value that summarizes the behavior of the function over that interval. It is calculated by taking the definite integral of the function over the interval and then dividing by the length of the interval.

• The formula for the average value of a function \(f(x)\) over \([a, b]\) is \( \frac{1}{b-a} \int_a^b f(x) \, dx \).
• This average provides insight into the overall level or trend of the function across the interval.

In our exercise, the average value was found to be \(\frac{14}{9}\) for \(f(x) = \sqrt{x+1}\) on the interval \([0,3]\). This average value allowed us to apply the Mean Value Theorem for Integrals, helping to pinpoint the exact location of \(c\) where \(f(c)\) equals this average.