Problem 8 - Calculus: Early Transcendentals

Question

Set up the integral to find the arc length of \(y=x e^{-x}\) on the interval [2,3] ; do not evaluate the integral. If you have access to appropriate software, approximate the value of the integral.

Step-by-Step Solution

Answer

The integral is \( \int_2^3 \sqrt{1 + e^{-2x}(1 - 2x + x^2)} \, dx. \)

Step 1: Understanding the Arc Length Formula

The arc length of a curve given by a function \( y = f(x) \) on an interval \([a, b]\) can be found using the formula:\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx. \] We need to set up this integral for the problem given.

Step 2: Calculate the Derivative

First, compute the derivative \( \frac{dy}{dx} \) for the function \( y = x e^{-x} \). Using the product rule, where \( u = x \) and \( v = e^{-x} \), we get: \( \frac{dy}{dx} = \frac{d}{dx}(x) e^{-x} + x \frac{d}{dx}(e^{-x}) = e^{-x} - x e^{-x}. \)

Step 3: Simplifying the Derivative

The derivative \( \frac{dy}{dx} = e^{-x} - x e^{-x} \) can be factored as \( \frac{dy}{dx} = e^{-x}(1 - x) \). This simplification will make the subsequent steps easier.

Step 4: Substitute into the Arc Length Formula

Substitute \( \frac{dy}{dx} = e^{-x}(1 - x) \) into the arc length formula. Inside the square root, we need \( 1 + \left( \frac{dy}{dx} \right)^2 \), which simplifies to: \[ 1 + (e^{-x}(1-x))^2 = 1 + e^{-2x}(1 - 2x + x^2). \]

Step 5: Set Up the Integral

The integral to find the arc length of the function \( y = x e^{-x} \) from \( x = 2 \) to \( x = 3 \) is:\[ L = \int_2^3 \sqrt{1 + e^{-2x}(1 - 2x + x^2)} \, dx. \] We have now completed setting up the integral.

Key Concepts

Integral Calculus

Integral calculus plays a crucial role in finding the arc length of a curve, among other things like areas under curves, volumes of solids, etc. The core idea of integral calculus is to sum up infinitely small quantities to determine a whole. In arc length calculations, we use integration to accumulate tiny straight-line segments along the curve from one point to another.

When you have a curve described by a function, such as \( y = f(x) \), the goal is to determine the total distance along the curve between two points. However, this distance is not just a simple difference between coordinates, as the path is not straight. Instead, you use the arc length integral, stepping through each infinitesimally small piece of the curve and summing them up over the interval \([a, b]\).

Integral calculus is essential in this context because it allows us to consider these small pieces as they approach zero in length, in a process known as taking the limit. This approach provides the exact length of the curve.

Derivative of Functions

Derivatives measure how a function changes as its input changes. They provide the slope of the tangent line to the curve at any point. Derivatives are fundamental in calculus as they help find optimal values, rates of change, and more.

For finding the arc length, derivatives are crucial. You need the derivative \( \frac{dy}{dx} \) of your function \( y = f(x) \) because it determines how steeply the function rises or falls. This slope is embedded in the arc length formula, where \( \left( \frac{dy}{dx} \right)^2 \) contributes to the overall integrand.

In our example with \( y = x e^{-x} \), the derivative using the product rule becomes \( \frac{dy}{dx} = e^{-x} - x e^{-x} \). Understanding this step ensures you know how the change in \( y \) relates to changes in \( x \) across the interval for accurate arc length calculations.

Arc Length Formula

The arc length formula is an elegant mathematical tool that provides the means to calculate the length of a curve in the Cartesian plane. Given a curve \( y = f(x) \) from \( x = a \) to \( x = b \), the arc length \( L \) is given by:

• \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \)

• This formula combines integration with the derivative of the function to account for both the linear distance and the curvature of the path.
• The term \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \) stems from Pythagoras' theorem, where you're summing the change along the x-axis and y-axis.

When setting up this integral for any specific function, it's important to properly simplify the expression under the square root. For instance, in \( y = x e^{-x} \), this simplification becomes \( 1 + e^{-2x}(1 - 2x + x^2) \), which makes the integration process more tractable.

Product Rule in Calculus

The product rule is a method used in calculus to differentiate functions that are products of two other functions. It's crucial for correctly finding derivatives when multiple functions are involved.

• According to the product rule, if you have a function \( y = u(x) v(x) \), the derivative \( \frac{dy}{dx} \) is given as: \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \).
• This means you take the derivative of the first function \( u(x) \) and multiply it by the second function, plus the first function multiplied by the derivative of the second function \( v(x) \).
• In our exercise, using the product rule, we find the derivative \( \frac{dy}{dx} = e^{-x} - x e^{-x} \) where the two functions are \( u(x) = x \) and \( v(x) = e^{-x} \).

Utilizing the product rule simplifies the process of finding derivatives for more complex expressions, which are then used in calculating the arc length and in many other applications of calculus.