Problem 23 - Calculus

Question

Find the function \(v(x)\) whose average value between 0 and \(x\) is \(\cos x .\) Start from \(\int_{0}^{x} v(t) d t=x \cos x\).

Step-by-Step Solution

Answer

The function is \(v(x) = \cos x - x \sin x\).

Step 1: Understand the Problem

We're given that the average value of the function \(v(t)\) over the interval from 0 to \(x\) is \(\cos x\). Mathematically, the average value of a function \(v(t)\) from \(a\) to \(b\) is given by: \[\frac{1}{b-a} \int_{a}^{b} v(t) \, dt.\] Since we're averaging from 0 to \(x\), \(b = x\) and \(a = 0\); so, \(b-a = x\). We also know the average is \(\cos x\). Thus, we have: \[\frac{1}{x} \int_{0}^{x} v(t) \, dt = \cos x.\] Given \(\int_{0}^{x} v(t) \, dt = x \cos x\), we aim to find \(v(x)\).

Step 2: Differentiate with Respect to \(x\)

To find \(v(x)\), we'll differentiate both sides of the equation \(\int_{0}^{x} v(t) \, dt = x \cos x\) with respect to \(x\). The left side becomes \(v(x)\) when using the Fundamental Theorem of Calculus: \[v(x) = \frac{d}{dx} \left( \int_{0}^{x} v(t) \, dt \right).\] The right side uses the product rule: \(\frac{d}{dx}(x \cos x) = \cos x - x \sin x\).

Step 3: Resulting Function

By differentiating, we have: \[v(x) = \cos x - x \sin x.\] This is the function \(v(x)\) whose average value over \([0, x]\) matches the given \(\cos x\).

Key Concepts

Average Value of a Function

The average value of a function over a certain interval gives us a sense of the function's central tendency. For any continuous function \( v(t) \) defined on a closed interval \([a, b]\), the formula for the average value is expressed as:

• \( \frac{1}{b-a} \int_{a}^{b} v(t) \, dt \)

This computes the total area under the curve of \( v(t) \) from \( a \) to \( b \) and then divides it by the length of the interval, \( b-a \).

In our problem, we seek the function \( v(x) \) whose average from \( 0 \) to \( x \) equals \( \cos x \). Hence, we apply the formula setting \( a = 0 \) and \( b = x \), yielding:

• \( \frac{1}{x} \int_{0}^{x} v(t) \, dt = \cos x \)

This means the integral of \( v(t) \) divided by \( x \) results in \( \cos x \). The challenge is to reverse engineer the integral to find the specific form of \( v(x) \).

Product Rule

The product rule is key for differentiating expressions where two functions are multiplied. It's a fundamental tool in calculus, helping us understand how to differentiate products of functions.

When faced with a product of two functions, say \( u(x) \times w(x) \), the derivative is calculated using:

• \( \frac{d}{dx}[u(x) \cdot w(x)] = u'(x) \cdot w(x) + u(x) \cdot w'(x) \)

This rule saves us from the error of just differentiating both functions individually, which wouldn't account for their interaction.

In solving for \( v(x) \), we see this applied in differentiating \( x \cos x \). Breaking it down:

• First function \( u(x) = x \), third function \( w(x) = \cos x \)
• The derivative becomes: \( \frac{d}{dx}(x \cos x) = 1 \cdot \cos x + x \cdot (-\sin x) \)

Resulting in the expression \( \cos x - x \sin x \), pivotal to solving our problem.

Differentiation

Differentiation is the process of finding the derivative of a function, which tells us the rate at which the function is changing at any point.

The Fundamental Theorem of Calculus aids in finding the derivative of an integral effectively. For a function \( F(x) = \int_{a}^{x} f(t) \, dt \), its derivative with respect to \( x \) is:

• \( F'(x) = f(x) \).

This principle allows us to differentiate integrals with ease and is used in our problem to identify \( v(x) \).

In practice, starting with the equation \( \int_{0}^{x} v(t) \, dt = x \cos x \), we apply differentiation on both sides:

• The left side becomes: \( v(x) \) directly, as per the theorem.
• The right side, already differentiated using the product rule, is: \( \cos x - x \sin x \).

Thus, differentiation reveals that the solution to our function \( v(x) \) over the interval \([0, x]\) is exactly \( \cos x - x \sin x \). This clear interpretation highlights how vital differentiation is to solve integration and calculus problems.