Problem 41 - Calculus

Question

If \(v(x)=|x-2|\) find \(f(x)\), Compute \(\int_{0}^{5} v(x) d x\).

Step-by-Step Solution

Answer

The integral is \(\frac{17}{2}\).

Step 1: Understand the Problem

We need to compute the integral \( \int_{0}^{5} v(x) \, dx \), where \( v(x) = |x - 2| \), from 0 to 5. This involves calculating the area under the curve of \( v(x) \) over the interval \([0, 5]\).

Step 2: Analyze the Function

The function \( v(x) = |x - 2| \) changes its expression depending on the value of \( x \). For \( x < 2 \), \( v(x) = 2 - x \). For \( x \geq 2 \), \( v(x) = x - 2 \). This implies a change at \( x = 2 \).

Step 3: Set Up the Integral

Split the integral at the point of change (\( x = 2 \)) to handle the absolute value function. Thus, we have two separate integrals: \( \int_{0}^{2} (2 - x) \, dx \) and \( \int_{2}^{5} (x - 2) \, dx \).

Step 4: Compute the First Integral

Compute \( \int_{0}^{2} (2 - x) \, dx \):\[ \int_{0}^{2} (2 - x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_0^2 = \left( 4 - 2 \right) - 0 = 2. \]

Step 5: Compute the Second Integral

Compute \( \int_{2}^{5} (x - 2) \, dx \):\[ \int_{2}^{5} (x - 2) \, dx = \left[ \frac{x^2}{2} - 2x \right]_2^5 = \left( \frac{25}{2} - 10 \right) - \left( 2 - 4 \right) = \frac{9}{2} + 2 = \frac{13}{2}. \]

Step 6: Combine the Results

Add the results of the two integrals to find the total area: \( 2 + \frac{13}{2} = \frac{4}{2} + \frac{13}{2} = \frac{17}{2}. \)

Key Concepts

Definite Integral

The concept of a definite integral is crucial when working with functions like the absolute value function. A definite integral allows us to compute the exact area under a curve over a specific interval on the x-axis. In other words, it helps find the total accumulation of quantities, such as the area in this case, between two bounds.

The definite integral is represented as \ \ \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively. Calculating a definite integral involves finding the antiderivative of the function, which then allows you to evaluate it at these two points. The result is the difference between these evaluations and represents the net area from \( a \) to \( b \).

In the exercise,"the definite integral \( \int_{0}^{5} \|x-2\| \, dx \)" was calculated, where the function needs to be handled carefully due to its piecewise nature."

Piecewise Functions

Piecewise functions are functions that have different expressions based on the value of the input \( x \). They can represent scenarios where a function behaves differently in different parts of its domain. This can occur in absolute value functions, such as \( v(x) = |x - 2| \).

For the interval considered, this function splits into two different linear expressions:

• \( v(x) = 2 - x \) when \( x < 2 \)
• \( v(x) = x - 2 \) when \( x \geq 2 \)

By breaking the function into pieces, we can analyze and integrate each portion separately.

Understanding the point of change in the function, \( x = 2 \), is essential for setting up the integral properly since it indicates where the function expression alters, creating distinct parts which need individual attention through separate integrals.

Area under a Curve

Calculating the area under a curve is a common goal when working with integrals. The area gives insight into the behavior and characteristics of the function over a specified interval.

When dealing with continuous functions, especially those that change form at certain points, such as piecewise functions, it's important to split the area calculation into manageable sections. For \( v(x) = |x - 2| \), evaluating the total area under the curve from \( x = 0 \) to \( x = 5 \) involved calculating two separate areas due to the expression change at \( x = 2 \).

• For \( x < 2 \), calculate the area using \( \int_{0}^{2} (2 - x) \, dx \).
• For \( x \geq 2 \), calculate the area using \( \int_{2}^{5} (x - 2) \, dx \).

The sum of these areas results in a total area. In the exercise, the total area calculated was \( \frac{17}{2} \), which represents the space enclosed by the curve \( v(x) \) and the x-axis over the given interval. This concept is vital in various fields such as physics, engineering, and economics, where areas under curves can translate to meaningful real-world phenomena.