Problem 28 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{0}^{2}|2 x-1| d x\)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{5}{2} \).

Step 1: Identify the Breakpoint of the Absolute Value

The given integral is \( \int_{0}^{2} |2x - 1| dx \). First, determine where the expression inside the absolute value, \( 2x - 1 \), equals zero. Solve \( 2x - 1 = 0 \) to find \( x = \frac{1}{2} \). This is the breakpoint at which the expression changes sign.

Step 2: Split the Integral at the Breakpoint

Since the absolute value function changes at \( x = \frac{1}{2} \), split the integral into two parts: \( \int_{0}^{\frac{1}{2}} |2x - 1| dx + \int_{\frac{1}{2}}^{2} |2x - 1| dx \). Analyze each part separately, considering the sign of \( 2x - 1 \) in each interval.

Step 3: Evaluate the Integral on the First Interval

For \( 0 \leq x < \frac{1}{2} \), the expression \( 2x - 1 \) is negative, so \( |2x - 1| = -(2x - 1) = 1 - 2x \). Thus, the first integral becomes \( \int_{0}^{\frac{1}{2}} (1 - 2x) dx \). Calculate this integral by finding the antiderivative: \[ \int (1 - 2x) \, dx = x - x^2 \]. Evaluate it from 0 to \( \frac{1}{2} \): \[ \left[ x - x^2 \right]_0^{\frac{1}{2}} = \left( \frac{1}{2} - \left(\frac{1}{2}\right)^2 \right) - (0 - 0) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \].

Step 4: Evaluate the Integral on the Second Interval

For \( \frac{1}{2} \leq x \leq 2 \), the expression \( 2x - 1 \) is positive, so \( |2x - 1| = 2x - 1 \). The second integral becomes \( \int_{\frac{1}{2}}^{2} (2x - 1) dx \). Calculate this integral by finding the antiderivative: \[ \int (2x - 1) \, dx = x^2 - x \]. Evaluate it from \( \frac{1}{2} \) to 2: \[ \left[ x^2 - x \right]_{\frac{1}{2}}^{2} = (4 - 2) - \left( \left(\frac{1}{2}\right)^2 - \frac{1}{2} \right) = 2 - \left( \frac{1}{4} - \frac{1}{2} \right) = 2 + \frac{1}{4} = \frac{9}{4} \].

Step 5: Combine the Results from Both Intervals

Add the results from the two integrals: \( \frac{1}{4} + \frac{9}{4} = \frac{10}{4} \). Simplify the fraction to get the final result: \( \frac{10}{4} = \frac{5}{2} \).

Key Concepts

Definite Integrals

Definite integrals might seem complex, but they are a fundamental concept in calculus. A definite integral is often used to calculate the area under a curve, and it has upper and lower limits which indicate the interval over which you’re integrating.
The notation \( \int_{a}^{b} f(x) \, dx \) asks you to find the area under the curve \( f(x) \) from \( x = a \) to \( x = b \).
In simpler terms:

• \( a \): your starting point on the \( x \)-axis.
• \( b \): your ending point on the \( x \)-axis.

To evaluate, find the antiderivative of \( f(x) \) and then use the Fundamental Theorem of Calculus, which states: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \] where \( F(x) \) is an antiderivative of \( f(x) \). Understanding definite integrals will help when dealing with areas and total quantities in physics and engineering.

Absolute Value Functions

Absolute value functions add an extra level of interest to calculus problems. The absolute value of a number is its distance from zero on the number line, without regard to direction.
Mathematically, it is noted as \( |x| \), and its piece-wise definition is:

• \( x \) if \( x \geq 0 \)
• \(-x \) if \( x < 0 \)

When you integrate an absolute value function like \( |2x - 1| \), it changes behavior at the point where the expression inside the absolute value is zero.
In our exercise, \( 2x - 1 = 0 \) gives us the breakpoint \( x = \frac{1}{2} \). Thus, the problem requires splitting the integral at this point to handle each segment appropriately. This ensures that you correctly account for how the function flips between positive and negative values over the defined interval.

Antiderivatives

Antiderivatives are the building blocks of integral calculus. They are functions that reverse the process of differentiation.
In other words, if you have a function \( f(x) \), an antiderivative \( F(x) \) is a function such that \( F'(x) = f(x) \).
Finding an antiderivative is often your first step in solving an integral. In practice:

• For \( f(x) = 1 - 2x \), an antiderivative is \( F(x) = x - x^2 \).
• For \( f(x) = 2x - 1 \), an antiderivative is \( F(x) = x^2 - x \).

Using these, you can compute definite integrals by evaluating the antiderivative at the boundaries of the integral. Always remember, there isn’t just one antiderivative but a family of them differing by a constant. However, for definite integrals, this constant cancels itself out.

Mathematical Integration Techniques

There are various integration techniques available to tackle different forms of integrals. Choosing the correct technique depends largely on the function you're dealing with. For functions involving absolute values, splitting the integral at points where changes occur is crucial.
Here are some common techniques you need to be aware of:

• Basic antiderivative technique: Used for simple polynomials or basic functions.
• Integration by parts: Useful for products of functions.
• Substitution method: Similar to reverse chain rule, handy when dealing with composite functions.

In our exercise, dealing with \(|2x - 1|\), we employed the technique of splitting the integral to accommodate the shift in the expression’s sign across the interval. Understanding when and how to apply each method makes solving integrals significantly more manageable and less daunting.