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Chapter 5: Problem 31

Find a positive value of \(k\) such that the average value of \(f(x)=\sqrt{3 x}\) over the interval \([0, k]\) is \(6 .\)

Short Answer

Expert verified
The positive value of \(k\) is 3.

Step by step solution

01

Understand the Problem

We need to find a positive value of \(k\) such that the average value of the function \(f(x) = \sqrt{3x}\) over the interval \([0, k]\) is 6.
02

Set Up the Average Value Formula

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by \(\frac{1}{b-a} \int_{a}^{b} f(x) \, dx\). In our case, \(a = 0\) and \(b = k\).
03

Substitute and Integrate

Substitute the function into the formula: \(\frac{1}{k-0} \int_{0}^{k} \sqrt{3x} \, dx = 6\). Integrate \(\sqrt{3x}\) from 0 to \(k\):\[\int_{0}^{k} \sqrt{3x} \, dx = \int_{0}^{k} (3x)^{0.5} \, dx = \frac{2}{3} (3x)^{1.5} \Big|_{0}^{k}.\]
04

Evaluate the Integral

Evaluate the definite integral:\[\frac{2}{3} (3x)^{1.5} \Big|_{0}^{k} = \frac{2}{3} \cdot \left[(3k)^{1.5} - (0)^{1.5} \right] = \frac{2}{3} \cdot (3k)^{1.5}.\]
05

Solve for k

The equation from Step 3 is \(\frac{1}{k} \cdot \frac{2}{3} (3k)^{1.5} = 6\). Simplify:\[\frac{2}{3} \cdot \frac{(3k)^{1.5}}{k} = 6.\]Resolving gives \(2 (3k)^{0.5} = 6\). Thus, \[(3k)^{0.5} = 3,\]so \[3k = 9.\]Divide by 3, \[k = 3.\]
06

Conclusion

The positive value of \(k\) that satisfies the condition is 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
The concept of a definite integral is essential in calculus as it helps us find the accumulation of quantities over a specific interval. When we take a definite integral of a function, we're essentially summing up small sections or slices of the function's graph to find the total "area" between the graph and the x-axis over an interval.
  • Integral Notation: The notation \( \int_{a}^{b} f(x) \, dx \) represents the definite integral of a function \( f(x) \) from \( a \) to \( b \).
  • Physical Interpretation: The result of a definite integral can often be interpreted as a physical quantity, such as area, volume, or total distance.
  • Evaluation: To compute a definite integral, we find the antiderivative (or the indefinite integral) of the function and then apply the limits of integration \( a \) and \( b \).
A simple yet powerful way to visualize the definite integral is to imagine how the function's graph changes between two points, adding up all the tiny bits of change. This becomes especially important when calculating things like the average value of a function over an interval, as seen in the original exercise.
Calculus Integration
Integration is a cornerstone of calculus that involves finding antiderivatives or summing up continuous distributions. The purpose of integration is to reverse the process of differentiation. While differentiation breaks down a curve into its rate of change, integration pieces it back together.
  • Antiderivative: An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that \( F'(x) = f(x) \). When integrated over an interval, this gives us the total change accumulated.
  • Techniques: Some common techniques of integration include substitution and integration by parts, which help tackle more complex functions.
  • Problem Solving: Integration allows us to solve a variety of real-world problems, from calculating areas and volumes to determining average values of functions as demonstrated.
Understanding integration and its applications can transform how we view and solve mathematical problems. It not only unifies various branches of calculus but also connects theory with practical, real-world scenarios.
Function Evaluation
Function evaluation is the process of determining what output a function gives for a particular input. This is fundamental in determining values for specific functions or using them within other mathematical operations.
  • Substitution: To evaluate a function, you replace the variable \( x \) with a given value. For instance, evaluating \( f(x) = \sqrt{3x} \) at \( x = 4 \) yields \( f(4) = \sqrt{12} \).
  • Intervals: When looking at intervals, like \([0, k]\), you evaluate the function not just at isolated points but across the entire range to understand its behavior over that stretch.
  • Applications: Function evaluations are crucial when setting up and solving integral equations, as seen in determining the average value of \( \sqrt{3x} \) over an interval, leading us to solve for \( k \).
By mastering function evaluation, students can better navigate through calculus problems, ensuring they comprehend how inputs relate to their respective outputs. This skill is foundational to calculus integration and solving problems involving definite integrals.

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Most popular questions from this chapter

(a) Prove that if \(f\) is an odd function, then $$ \int_{-a}^{a} f(x) d x=0 $$ and give a geometric explanation of this result. [Hint: One way to prove that a quantity \(q\) is zero is to show that \(q=-q .]\) (b) Prove that if \(f\) is an even function, then $$ \int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x $$ and give a geometric explanation of this result. [Hint: Split the interval of integration from \(-a\) to \(a\) into two parts at \(0 .]\)

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