Math Problem Statement
Karjakin
Solution
To solve question 17, we need to calculate the integral of the piecewise function from to using appropriate geometric area formulas. The function is defined as follows:
x & \text{if } 0 \leq x < 1 \\ 1 & \text{if } 1 \leq x \leq 3 \\ x - 4 & \text{if } 3 < x \leq 5 \end{cases}$$ ### Step-by-Step Solution To evaluate $$\int_0^5 f(x) \, dx$$, we can split the integral based on the intervals defined by the piecewise function. #### 1. Integral from $$x = 0$$ to $$x = 1$$: On this interval, $$f(x) = x$$. $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$$ #### 2. Integral from $$x = 1$$ to $$x = 3$$: On this interval, $$f(x) = 1$$. $$\int_1^3 1 \, dx = \left[ x \right]_1^3 = 3 - 1 = 2$$ #### 3. Integral from $$x = 3$$ to $$x = 5$$: On this interval, $$f(x) = x - 4$$. $$\int_3^5 (x - 4) \, dx = \left[ \frac{x^2}{2} - 4x \right]_3^5$$ Calculating this: 1. Substitute $$x = 5$$: $$\frac{5^2}{2} - 4 \times 5 = \frac{25}{2} - 20 = \frac{25 - 40}{2} = \frac{-15}{2} = -7.5$$ 2. Substitute $$x = 3$$: $$\frac{3^2}{2} - 4 \times 3 = \frac{9}{2} - 12 = \frac{9 - 24}{2} = \frac{-15}{2} = -7.5$$ The integral over this interval is: $$-7.5 - (-7.5) = -7.5 + 7.5 = 0$$ ### Combining Results Now, we add up the results from each interval: $$\int_0^5 f(x) \, dx = \frac{1}{2} + 2 + 0 = \frac{1}{2} + 2 = 2.5$$ ### Final Answer $$\int_0^5 f(x) \, dx = 2.5$$ --- Would you like further details on this solution or have additional questions? Here are some related questions for further exploration: 1. How would the integral change if the interval was from $$x = 0$$ to $$x = 4$$? 2. Can we visualize the graph of $$f(x)$$ and interpret each integral as an area under the curve? 3. How does the piecewise nature of $$f(x)$$ affect the computation of the integral? 4. What is the interpretation of each interval’s contribution to the total area? 5. How would we approach integrating a more complex piecewise function with more intervals? **Tip:** For piecewise functions, break the integral into separate parts according to each interval and evaluate each segment individually before summing the results.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Integral Calculus
Piecewise Functions
Definite Integrals
Area Calculation
Formulas
Definite integral \( \int_a^b f(x) \, dx \)
Integral of linear functions \( \int x \, dx = \frac{x^2}{2} + C \)
Area under constant functions \( \int a \, dx = a(x - x_0) \)
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
Grades 11-12
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