Problem 49 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the definite integral by the most convenient method. Explain your approach. $$ \int_{0}^{4}(2-|x-2|) d x $$

Step-by-Step Solution

Answer

The definite integral \(\int_{0}^{4}(2-|x-2|)dx\) equals 6

Step 1: Divide into two integrals

For \(x \leq 2\) the absolute value function will be equal to \(2 - x\), while for \(x > 2\) it will be equal to \(x - 2\). The integral can hence be rewritten as a sum of two integrals:\[\int_{0}^{4}(2-|x-2|) dx = \int_{0}^{2}(2 - x) dx + \int_{2}^{4}(x - 2) dx\]

Step 2: Evaluate the first integral

Evaluating the first of these integrals gives:\[\int_{0}^{2} (2 - x) dx = \left[2x-\frac{1}{2} x^2\right]_{0}^{2} = 2*2-\frac{1}{2}*2^2 - (2*0-\frac{1}{2}*0^2) = 4 - 2 = 2\]

Step 3: Evaluate the second integral

Evaluating the second integral gives:\[\int_{2}^{4} (x - 2) dx = \left[\frac{1}{2}x^2 - 2x\right]_{2}^{4} = \frac{1}{2}*4^2-2*4 - (\frac{1}{2}*2^2-2*2) = 8 - 4 = 4\]

Step 4: Sum the calculated areas

The total area under the graph from \(x = 0\) to \(x = 4\) is the sum of the areas calculated in the previous steps:\[\int_{0}^{4}(2-|x-2|) dx = 2 + 4 = 6\]

Key Concepts

Calculus

Calculus is a branch of mathematics that involves studying how things change. It helps us understand the rate of change of quantities and allows us to compute the area under curves. The two primary branches of calculus are differential calculus and integral calculus.
In the context of our exercise, we focus on integral calculus, specifically, definite integrals. This helps us calculate the total area under a curve between two points. Calculus becomes handy in many real-world applications such as physics, engineering, and economics. Through calculus, we are able to understand and explain dynamic systems, allowing for precise calculations that inform decision making.
We used definite integrals in the exercise, computing the area under the curve represented by \(2-|x-2|\) from \(x=0\) to \(x=4\). By breaking down the function and calculating step by step, calculus lets us find numeric solutions to complex problems.

Absolute Value Function

An absolute value function is represented by \(|x|\), which gives the distance of a number from zero on the number line, ensuring all outputs are non-negative. In this exercise, the expression \(|x-2|\) forces us to consider two cases based on the value of \(x\). This is because absolute values change the behavior of the function depending on whether the expression inside is negative or positive.
- When \(x \leq 2\), \(|x-2|\) simplifies to \(2-x\).
- When \(x > 2\), \(|x-2|\) simplifies to \(x-2\).
By understanding how absolute value functions work, one can correctly split the integral into manageable parts based on these conditions. This step is crucial as it impacts how we compute the definite integral by ensuring our function behaves correctly over its domain.

Integration by Parts

Integration by parts is a technique used to solve integrals that involve products of functions. Although not directly used in our specific exercise, it is an essential tool in calculus for handling integrals that cannot be simplified easily using basic rules.
To use integration by parts, apply the formula \(\int u \, dv = uv - \int v \, du\), where selecting \(u\) and \(dv\) appropriately helps in transforming a complex integral into simpler parts. This technique is useful when traditional methods like substitution or basic rule applications result in complicated forms. By breaking down an integral into smaller, manageable components, it simplifies more challenging integrations.
Though the definite integral in our exercise was relatively straightforward, understanding integration by parts prepares you for tackling more complex integrals that you might encounter in different calculus problems.

Area Under a Curve

Calculating the area under a curve is one of the primary applications of definite integrals in calculus. The definite integral \(\int_{a}^{b} f(x) \, dx\) represents the signed area between the curve \(y = f(x)\) and the x-axis from \(x = a\) to \(x = b\).
In our exercise, finding the area under the curve \(2-|x-2|\) from 0 to 4 is solved by breaking it into two areas:
- The area from \(x=0\) to \(x=2\) where the function \(2-x\) provides the curve.
- The area from \(x=2\) to \(x=4\) where the function \(x-2\) creates the curve.
Adding these computed areas gives the total area under the curve.
Understanding the area under a curve concept is vital because it not only applies to geometrical problems but also extends to physics and other fields, representing quantities such as total displacement or accumulated differences.