Problem 31 - Applied Calculus

Question

Compute the definite integral and interpret the result in terms of areas. $$\int_{1}^{4}(x-3 \ln x) d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{33}{2} - 12 \ln 4 \), which is the net area. This indicates more area is above the x-axis than below on the interval \([1, 4]\).

Step 1: Identify the function and interval

The function we need to integrate is \( f(x) = x - 3 \ln x \) over the interval from \( x = 1 \) to \( x = 4 \). This integral will give us the net area between the function and the x-axis over the interval.

Step 2: Compute the antiderivative

To compute the definite integral, we first need to find the antiderivative \( F(x) \) of the function \( f(x) = x - 3 \ln x \). - The antiderivative of \( x \) is \( \frac{x^2}{2} \).- The antiderivative of \( 3 \ln x \) is \( 3(x \ln x - x) \), by using integration by parts.Thus, the antiderivative \( F(x) \) is:\[ F(x) = \frac{x^2}{2} - 3(x \ln x - x) \] \[ = \frac{x^2}{2} - 3x \ln x + 3x \].

Step 3: Evaluate the antiderivative at the bounds

We now substitute the upper and lower bounds of the interval into the antiderivative:- \( F(4) = \frac{4^2}{2} - 3(4 \ln 4 - 4) = 8 - 3(4 \ln 4 - 4) \). - \( F(1) = \frac{1^2}{2} - 3(1 \ln 1 - 1) = \frac{1}{2} + 3 \).Simplifying gives us:- \( F(4) = 8 - 12 \ln 4 + 12 = 20 - 12 \ln 4 \).- \( F(1) = \frac{1}{2} + 3 = \frac{7}{2} \).

Step 4: Compute the definite integral

Calculate the definite integral by subtracting these values:\[ \int_{1}^{4}(x-3 \ln x) dx = F(4) - F(1) \]\[ = (20 - 12 \ln 4) - \frac{7}{2} \]\[ = 20 - 12 \ln 4 - \frac{7}{2} \]\[ = \frac{40}{2} - \frac{7}{2} - 12 \ln 4 \]\[ = \frac{33}{2} - 12 \ln 4 \].

Step 5: Interpret in terms of areas

The result \( \frac{33}{2} - 12 \ln 4 \) represents the net area between the curve \( y = x - 3 \ln x \) and the x-axis from \( x = 1 \) to \( x = 4 \). Since this value is not negative, it means that the area above the x-axis exceeds the area below the x-axis over the interval.

Key Concepts

Antiderivative

The antiderivative, often called the indefinite integral, of a function is a fundamental concept in calculus. It refers to a function whose derivative is the original function given. The process of finding an antiderivative is termed "integration".
To solve a definite integral such as \( \int_{1}^{4}(x-3 \ln x) dx \), you must first determine the antiderivative of the function \( f(x) = x - 3 \ln x \).
This involves recognizing individual parts:

• The antiderivative of \( x \) is \( \frac{x^2}{2} \), as differentiating \( \frac{x^2}{2} \) gives back \( x \).
• For the \( 3 \ln x \) component, integration by parts is necessary, a technique to handle products of functions, which will be detailed in another section.

The complete antiderivative is formulated as:\[ F(x) = \frac{x^2}{2} - 3(x \ln x - x). \]

Net Area

In calculus, the concept of net area is pivotal when interpreting definite integrals. The definite integral \( \int_{a}^{b} f(x) dx \) represents the net area between the function \( y = f(x) \) and the x-axis over an interval \[ [a, b] \].
This encompasses:

• The area above the x-axis, after subtracting the area below it.
• If the outcome is positive, more area lies above the x-axis within the bounds \( a \) to \( b \).

In our exercise, calculating the definite integral \( \int_{1}^{4}(x-3 \ln x) dx \) yields \( \frac{33}{2} - 12 \ln 4 \), signifying the net area from \( x = 1 \) to \( x = 4 \), indicating a predominance of area above the x-axis as it is a non-negative value.

Integration by Parts

One important technique for finding antiderivatives is integration by parts, especially useful when dealing with products of functions like polynomials and logarithms.
The formula for integration by parts is derived from the product rule for differentiation: \[ \int u \; dv = uv - \int v \; du \].
Steps include:

• Identifying parts of the expression as \( u \) and \( dv \).
• Calculating \( du \) and \( v \) through differentiation and integration respectively.
• Applying the formula to solve for the integral.

In the exercise, for the part \( 3 \ln x \), you can set \( u = \ln x \) (therefore \( du = \frac{1}{x} dx \)) and \( dv = 3 dx \) (therefore \( v = 3x \)). Executing these steps assists in solving components of the integral that aren't easily integrated through simpler methods alone.