Problem 53 - Calculus Single Variable

Question

For what value of \(c\) is \(-1 / 2\) the average value of \((x-c)\) \(\sin (x)\) over the interval \([0, \pi / 3] ?\)

Step-by-Step Solution

Answer

The value of \(c\) is \(-\frac{\pi}{2} + \frac{3\sqrt{3}}{4}\).

Step 1: Understanding Average Value Formula

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by the formula:\[\text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]In this problem, we need to find the average value of \((x-c) \sin(x)\) over the interval \([0, \pi/3]\).

Step 2: Setup the Integral

In the formula above, substitute \(f(x) = (x-c) \sin(x)\), \(a = 0\), and \(b = \pi/3\). The average value is given as \(-1/2\). Set up the equation as:\[\frac{1}{\pi/3 - 0} \int_{0}^{\pi/3} (x-c) \sin(x) \, dx = -\frac{1}{2}\]This simplifies to:\[\frac{3}{\pi} \int_{0}^{\pi/3} (x-c) \sin(x) \, dx = -\frac{1}{2}\]

Step 3: Evaluate the Integral

Compute the integral \(\int_{0}^{\pi/3} (x-c) \sin(x) \, dx\). Using integration by parts, let \(u = x-c\) and \(dv = \sin(x) \, dx\). Then, \(du = dx\) and \(v = -\cos(x)\).The integration by parts formula is:\[\int u \, dv = uv - \int v \, du\]So,\[\int_{0}^{\pi/3} (x-c) \sin(x) \, dx = \left[-(x-c)\cos(x)\right]_{0}^{\pi/3} + \int_{0}^{\pi/3} \cos(x) \, dx\]

Step 4: Solve the Integrals in Parts

Evaluate the expression:1. Calculate \(\left[-(x-c)\cos(x)\right]_{0}^{\pi/3}\): - At \(x = \pi/3\): \(-(\pi/3-c)\cos(\pi/3) = -((\pi/3)-c)\left(\frac{1}{2}\right) = -\frac{(\pi/3-c)}{2}\) - At \(x = 0\): \(-(0-c)\cos(0) = -c\)2. Calculate \(\int_{0}^{\pi/3} \cos(x) \, dx\): - Find antiderivative: \(\sin(x)\) - Evaluate from 0 to \(\pi/3\): \(\sin(\pi/3) - \sin(0) = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2}\)Putting it all together:\[-\frac{(\pi/3-c)}{2} + c + \frac{\sqrt{3}}{2}\]

Step 5: Plug Back into the Average Value Equation

Equate the integral expression to the average value result:\[\frac{3}{\pi} \left[-\frac{(\pi/3-c)}{2} + c + \frac{\sqrt{3}}{2}\right] = -\frac{1}{2}\]Simplify and solve the context equation for \(c\).

Step 6: Solving for \(c\) Value

Distribute and rearrange the expression:\[\frac{3}{\pi} \left[-\frac{\pi}{6} + \frac{c}{2} + c + \frac{\sqrt{3}}{2}\right] = -\frac{1}{2}\]Simplify further to:\[\frac{3}{\pi} \left[\frac{2c - \pi}{6} + \frac{\sqrt{3}}{2}\right] = -\frac{1}{2}\]Solve for \(c\) upon rearrangement and simplification.

Key Concepts

Integration by Parts

When faced with integrals that are products of functions, a powerful tool we can use is Integration by Parts. This technique is like a trade-off. It allows us to exchange a difficult integral for something easier. The basic formula for Integration by Parts comes from the Product Rule for derivatives. It is expressed as:

• \[ \int u \, dv = uv - \int v \, du \]

In this scenario, we have two parts, namely 'u' and 'dv'. Choosing the right 'u' and 'dv' is crucial for simplifying the integral.For our problem, we set:

• \( u = x-c \)
• \( dv = \sin(x) \, dx \)

This gives us derivatives and integrals:

• \( du = dx \)
• \( v = -\cos(x) \)

Using the formula involves plugging these into the equation. This splits the original integral into two parts that we can evaluate separately.First, we compute the boundary terms \(-((x-c)\cos(x))\), then handle the remaining integral of \(\cos(x)\). This strategic method simplifies our task, making it manageable to solve complex products.

Definite Integrals

Definite integrals help us find the accumulation or net area under a curve; it's like measuring the space on a graph between two points. For any continuous function \(f(x)\) from \(a\) to \(b\), a definite integral \(\int_{a}^{b} f(x) \, dx\) represents this area.In this particular problem, we're working with the definite integral of the function \((x-c) \sin(x)\) over the interval \([0, \pi/3]\). This involves evaluating the function at the two endpoints and computing the integral of a product of functions.The process begins by using Integration by Parts to translate the integral into simpler parts. After this conversion, the definite integral provides values after substituting the bounds of the interval \(0\) and \(\pi/3\). This allows us to apply the Fundamental Theorem of Calculus: evaluate the antiderivative at \(a\) and \(b\) and then subtract. This results in a precise net area, helping solve the average value equation for \(c\).

Trigonometric Functions

Trigonometric functions like \(\sin(x)\) and \(\cos(x)\) are foundational in mathematics and appear often in integrals and various calculations. They have well-known properties that make them ideal for many applications, including solving definite integrals in our exercise.The function \(\sin(x)\) is periodic and oscillates between -1 and 1. Its antiderivative is \(-\cos(x)\), a key relationship used in integration by parts.When evaluating integrals, we frequently use these trigonometric identities to simplify our calculations. This includes resolving the integral of \(\cos(x)\) over the specified interval, which becomes:

• \[ \int_{0}^{\pi/3} \cos(x) \, dx = \sin(x) \Big|^{\pi/3}_{0} \]

Substituting the values gives us the specific result that further aids in evaluating our main average value expression.These trigonometric properties and their integrals combine seamlessly with algebraic manipulation to find potential solutions, like the average value, essential for our exercise's completion.