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Magazine Chapter 7: Problem 44
In each of Exercises \(43-52\) calculate the average of the given expression over the given interval. $$ x \sin (x) \quad 0 \leq x \leq \pi $$
Short Answer
Expert verified
The average value is 1.
Step by step solution
01
Understand the Average Value Formula
To find the average value of a function over a given interval \[ a, b \], we use the formula: \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] Here, the function is \( x \sin(x) \) and the interval is \( [0, \pi] \). Therefore, \( a = 0 \) and \( b = \pi \).
02
Set Up the Integral
Substitute the values of the function and the interval into the formula for the average value:\[ f_{avg} = \frac{1}{\pi - 0} \int_{0}^{\pi} x \sin(x) \, dx \]This simplifies to:\[ f_{avg} = \frac{1}{\pi} \int_{0}^{\pi} x \sin(x) \, dx \]
03
Integrate by Parts
The integral \( \int x \sin(x) \, dx \) requires integration by parts. Let \( u = x \) and \( dv = \sin(x) \, dx \). Then \( du = dx \) and \( v = -\cos(x) \). Using integration by parts:\[ \int x \sin(x) \, dx = -x \cos(x) \Big|_0^\pi + \int \cos(x) \, dx \]Solve \( \int \cos(x) \, dx \):\[ \int \cos(x) \, dx = \sin(x) \Big|_0^\pi \]
04
Evaluate the Integral
Substitute the known values back into the integral expression:\[ -x \cos(x) \Big|_0^\pi + \sin(x) \Big|_0^\pi = -\pi \cos(\pi) + 0 + (\sin(\pi) - \sin(0)) \]Evaluate:\[ = (-\pi (-1) + 0 + 0) = \pi \]
05
Calculate the Average Value
Substitute the result of the integral back into the average value formula:\[ f_{avg} = \frac{1}{\pi} \times \pi = 1 \]
06
Final Step: Write the Conclusion
The average value of the function \( x \sin(x) \) over the interval \( [0, \pi] \) is 1.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is an important technique in calculus used to solve integrals where standard integration methods may not work easily. It's based on the product rule for differentiation and is particularly useful for integrals involving a product of functions.
To perform integration by parts, you choose parts of the integral as:
\[ \int u \, dv = uv - \int v \, du \]This process allows you to transform a complex integral into simpler parts that might be easier to solve. In the given problem, the function x sin(x) is integrated by choosing u = x and dv = sin(x)dx.
To perform integration by parts, you choose parts of the integral as:
- Let one part be u, which is a function that is easily differentiable.
- Let another part be dv, which is a part of the integral that can be easily integrated.
- du is the derivative of u.
- v is the integral of dv.
\[ \int u \, dv = uv - \int v \, du \]This process allows you to transform a complex integral into simpler parts that might be easier to solve. In the given problem, the function x sin(x) is integrated by choosing u = x and dv = sin(x)dx.
Definite Integral
A definite integral is a fundamental concept in calculus that computes the accumulation of quantities between two bounds, represented as \[ \int_{a}^{b} f(x) \, dx \], where is the lower limit and is the upper limit. In practice, a definite integral can find the total area under the curve of a function between two limits.
Unlike indefinite integrals, which represent a family of functions, definite integrals yield a number, a specific accumulated value.
This is useful for calculating averages, total amounts of change, or in this case, the average value of the function over a given interval. The definite integral is widely used in physics and engineering to determine quantities like area, volume, displacement, and more.
In our problem with the function x sin(x), the definite integral calculation over <0> to <\pi> is crucial in finding the average value. The ultimate result is the substitution and evaluation of the limits to find the final value.
Unlike indefinite integrals, which represent a family of functions, definite integrals yield a number, a specific accumulated value.
This is useful for calculating averages, total amounts of change, or in this case, the average value of the function over a given interval. The definite integral is widely used in physics and engineering to determine quantities like area, volume, displacement, and more.
In our problem with the function x sin(x), the definite integral calculation over <0> to <\pi> is crucial in finding the average value. The ultimate result is the substitution and evaluation of the limits to find the final value.
Trigonometric Functions
Trigonometric functions are essential in calculus for representing periodic phenomena. Functions like sine, cosine, and tangent relate to angles and the unit circle, and they are foundational in the study of oscillatory behaviors.
The sine function, represented as sin(x), outputs the y-coordinate of a point on the unit circle at an angle x from the positive x-axis. This periodic nature of sine is crucial for many real-world applications, including waves and oscillations.
In the problem at hand, the use of sin(x) in the integral indicates a periodic behavior that needs to be considered during integration. In particular, knowing the derivatives and integrals of sin(x) helps solve the problem using integration by parts, as seen in the exercise solution.
The sine function, represented as sin(x), outputs the y-coordinate of a point on the unit circle at an angle x from the positive x-axis. This periodic nature of sine is crucial for many real-world applications, including waves and oscillations.
In the problem at hand, the use of sin(x) in the integral indicates a periodic behavior that needs to be considered during integration. In particular, knowing the derivatives and integrals of sin(x) helps solve the problem using integration by parts, as seen in the exercise solution.
Calculus
Calculus is the branch of mathematics focusing on continuous change, comprising two main branches: differential and integral calculus.
Differential calculus regards rates of change and slopes of curves, while integral calculus focuses on the accumulation of quantities and the areas under and between curves.
Both branches rely on the concept of a limit to define and solve problems involving changes. Calculus is indispensable across the sciences, from physics to economics, allowing precise calculation of dynamic systems.
In our problem, calculus is applied to determine the average function value over an interval. The exercise involves both integration (accumulation) to solve the integral part and algebraic manipulation to find the average which shows how calculus can solve real-world phenomena by analyzing changing quantities.
Differential calculus regards rates of change and slopes of curves, while integral calculus focuses on the accumulation of quantities and the areas under and between curves.
Both branches rely on the concept of a limit to define and solve problems involving changes. Calculus is indispensable across the sciences, from physics to economics, allowing precise calculation of dynamic systems.
In our problem, calculus is applied to determine the average function value over an interval. The exercise involves both integration (accumulation) to solve the integral part and algebraic manipulation to find the average which shows how calculus can solve real-world phenomena by analyzing changing quantities.
