Problem 48 - Calculus: Early Transcendental Functions

Question

Compute \(\int_{0}^{4} f(x) d x\). $$f(x)=\left\\{\begin{array}{ll} 2 & \text { if } x \leq 2 \\ 3 x & \text { if } x>2 \end{array}\right.$$

Step-by-Step Solution

Answer

The result of computing the integral of the function \(f(x)\) from 0 to 4 is 22.

Step 1: Compute the integral for x ≤ 2

In this step, we're computing the integral \(\int_{0}^{2} 2 d x\). This is a straightforward, definite integral, with the integrand being a constant(2). To compute this, you just multiply the integrand by the width of the integration interval(2-0). So, \(\int_{0}^{2} 2 d x = 2 * (2 - 0) = 4.\)

Step 2: Compute the integral for x > 2

In this step, we're computing \(\int_{2}^{4} 3x d x\). To evaluate this, we'll use the power rule for antiderivatives -- where \(\int x^n d x = (x^{n+1}) /(n+1)\). Applying this rule, we get [(3*4^2)/2] - [(3*2^2)/2] = (24-6)=18.

Step 3: Add the results

We now add the results from Step 1 and Step 2: \(4 + 18 = 22\). So the total integral \(\int_{0}^{4} f(x) d x = 22\).

Key Concepts

Understanding Piecewise Functions

In calculus, piecewise functions are functions that have different expressions or rules for different parts of their domain. This allows a single function to behave differently depending on the input value, which helps in modeling real-world scenarios that are not uniform. In the given exercise, we see a piecewise function:

• For values of \( x \) that are less than or equal to 2, the function is defined as \( f(x) = 2 \).
• For values of \( x \) greater than 2, the function is \( f(x) = 3x \).

Each part of the piecewise function must be integrated separately within its specific interval. This means that for our exercise, we first compute the integral from 0 to 2 for \( f(x) = 2 \) and then from 2 to 4 for \( f(x) = 3x \). Breaking down the problem this way ensures that we accurately capture the behavior of the function across its entire domain.

Applying the Power Rule

The power rule is a fundamental technique in calculus used to find antiderivatives and integrals of polynomial functions. It states that the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) plus a constant of integration, usually denoted by \( C \). However, when dealing with definite integrals, the constant cancels out since we are calculating the net area between bounds.In the exercise, we apply the power rule in Step 2 while integrating \( 3x \) from 2 to 4. Since \( x \) is to the first power,\( n \) equals 1 in this case. Using the power rule, we perform the following steps:

• Find the antiderivative of \( 3x \). This gives us \( \frac{3x^2}{2} \).
• Evaluate this expression between the upper limit 4 and the lower limit 2, which calculates to \( \left[ \frac{3 \cdot 4^2}{2} \right] - \left[ \frac{3 \cdot 2^2}{2} \right] = 24 - 6 = 18 \).

This method efficiently provides the area under \( f(x) = 3x \) from \( x = 2 \) to \( x = 4 \).

The Role of Antiderivatives in Definite Integrals

Antiderivatives, also known as indefinite integrals, are functions whose derivatives recreate the original function. They are crucial for calculating definite integrals, which represent the net area under a curve between two points on the x-axis.When calculating a definite integral, such as in this exercise, the fundamental technique involves finding the antiderivative of the function and then performing the following:

• Evaluate the antiderivative at the upper limit of the integration interval.
• Subtract the value of the antiderivative evaluated at the lower limit.

For the piecewise function, two separate antiderivatives are calculated due to the change in the function's rule at \( x = 2 \):

• The integral of the constant \( 2 \), which is simply multiplied by the interval width to give \( 4 \).
• The antiderivative of \( 3x \), using the power rule, yields 18 as shown in the calculation.

Adding these results gives the complete area under the curve from \( x = 0 \) to \( x = 4 \), totaling \( 22 \). Understanding antiderivatives simplifies the process of computing definite integrals.