Problem 17 - Calculus

Question

Calculate \(\int_{a}^{b} f(x) d x,\) where \(a\) and \(b\) are the left and right end points for which fis defined, by using the Interval Additive Property and the appropriate area formulas from plane geometry. Begin by graphing the given function. $$ f(x)=\left\\{\begin{array}{ll} 2 x & \text { if } 0 \leq x \leq 1 \\ 2 & \text { if } 1

Step-by-Step Solution

Answer

The integral \( \int_{0}^{5} f(x)\,dx = \frac{27}{2} \).

Step 1: Graph the Function

Let's start by sketching the piecewise function \( f(x) \). For \( 0 \leq x \leq 1 \), \( f(x) = 2x \), which is a line starting at \( (0, 0) \) and ending at \( (1, 2) \). For \( 1 < x \leq 2 \), \( f(x) = 2 \), which is a horizontal line from \( (1, 2) \) to \( (2, 2) \). Lastly, for \( 2 < x \leq 5 \), \( f(x) = x \), which is a line from \( (2, 2) \) to \( (5, 5) \).

Step 2: Identify the Intervals

The function is defined in different pieces, so we'll break the integral into three intervals: \([0, 1], [1, 2], \) and \([2, 5]\). We will calculate each part individually.

Step 3: Integrate Over [0, 1]

For \( 0 \leq x \leq 1 \), \( f(x) = 2x \). The integral is \( \int_{0}^{1} 2x\,dx \). This will involve using the power rule for integration.

Step 4: Calculate the Integral Over [0, 1]

To integrate \( 2x \), apply the power rule: \( \int 2x\,dx = x^2 + C \). Compute \( \left. x^2 \right|_0^1 = 1^2 - 0^2 = 1 \). So, \( \int_{0}^{1} 2x\,dx = 1 \).

Step 5: Integrate Over [1, 2]

For \( 1 < x \leq 2 \), \( f(x) = 2 \). The integral is \( \int_{1}^{2} 2\,dx \). This is a constant function, so just multiply the constant by the length of the interval.

Step 6: Calculate the Integral Over [1, 2]

To find \( \int_{1}^{2} 2\,dx \), compute \( 2 \cdot (2 - 1) = 2 \). So, \( \int_{1}^{2} 2\,dx = 2 \).

Step 7: Integrate Over [2, 5]

For \( 2 < x \leq 5 \), \( f(x) = x \). The integral is \( \int_{2}^{5} x\,dx \). Use the power rule for this integration.

Step 8: Calculate the Integral Over [2, 5]

To integrate \( x \), apply the power rule: \( \int x\,dx = \frac{x^2}{2} + C \). Compute \( \left. \frac{x^2}{2} \right|_2^5 = \left( \frac{25}{2} \right) - \left( \frac{4}{2} \right) = \frac{21}{2} \). So, \( \int_{2}^{5} x\,dx = \frac{21}{2} \).

Step 9: Add the Results

Combine the results of the integrals: \( 1 + 2 + \frac{21}{2} = 3 + \frac{21}{2} = \frac{6}{2} + \frac{21}{2} = \frac{27}{2} \).

Step 10: Final Conclusion

The integral \( \int_{0}^{5} f(x)\,dx = \frac{27}{2} \).

Key Concepts

Piecewise Functions

Piecewise functions are a unique type of function that have different expressions for different intervals within their domain. In simpler terms, this means that the function behaves differently depending on the value of the input variable. For example, in the given exercise, the function \( f(x) \) consists of three different expressions:

• For \( 0 \leq x \leq 1 \), the function is \( f(x) = 2x \).
• For \( 1 < x \leq 2 \), the function is a constant \( f(x) = 2 \).
• For \( 2 < x \leq 5 \), the function transforms to \( f(x) = x \).

This structure allows us to handle complex behaviors by breaking down the function into manageable parts, each with its own rule. When integrating piecewise functions, it is important to respect these intervals and compute the integral separately over each part, as we did in the solution.

Interval Additive Property

The interval additive property is a fundamental rule of integration that states the integral of a function over an interval can be calculated by dividing the interval into smaller sub-intervals and summing the integrals over these sub-intervals. In mathematical terms, for a function \( f(x) \) defined over an interval \([a, b]\), and any intermediate point \( c \), the property states:\[\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx\]This property is essential when working with piecewise functions like in our exercise, where the domain was divided into three separate intervals: \([0, 1]\), \([1, 2]\), and \([2, 5]\). Each interval was individually integrated to accommodate the changes in the function's definition, and finally, all results were added together to find the overall integral. This approach ensures that we precisely account for the function's behavior across different segments of the domain.

Area Formulas in Geometry

In the context of definite integration, especially when dealing with piecewise functions, understanding geometry and its area formulas can be very beneficial. Integration can often be visualized as finding the area under a curve for a given interval on the x-axis. Some areas might form simple geometric shapes such as rectangles or triangles.For instance, in the original exercise for the interval \([1,2]\), the function \( f(x) = 2 \) is constant, creating a rectangular region under the curve. The area, calculated by multiplying height by the width, easily translates to the integral value over that interval. Similarly, for linear segments such as \( f(x) = 2x \) in the interval \([0,1]\), the shape under the curve forms a right triangle. Here, the area can be computed using the formula \( \frac{1}{2} \times \text{base} \times \text{height} \). Understanding these geometric interpretations can simplify the process of integration and provide insights into why certain calculations yield specific results. Utilizing area formulas effectively allows students to visually understand the integrals, enhancing their comprehension of the relationship between definite integration and geometry.