Math Problem Statement
solve this problem
The function f is defined above. The value of \( \int_3^{-5} f(x) \, dx \) is
Solution
To solve this problem, let's analyze the function and the integral provided.
The function is defined as:
\begin{cases} \frac{|x|}{x} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$$ ### Step 1: Simplify $$f(x)$$ for $$x \neq 0$$ For $$x \neq 0$$: - When $$x > 0$$, $$|x| = x$$, so $$f(x) = \frac{x}{x} = 1$$. - When $$x < 0$$, $$|x| = -x$$, so $$f(x) = \frac{-x}{x} = -1$$. Thus, $$f(x) = \begin{cases} 1 & \text{for } x > 0 \\ -1 & \text{for } x < 0 \end{cases}$$ ### Step 2: Set up the Integral We are asked to find the value of: $$\int_3^{-5} f(x) \, dx$$ Notice that the integral is from $$3$$ to $$-5$$, which means we are integrating in the opposite direction of the standard $$a$$ to $$b$$ notation (where $$a < b$$). We can reverse the limits and introduce a negative sign: $$\int_3^{-5} f(x) \, dx = -\int_{-5}^3 f(x) \, dx$$ ### Step 3: Split the Integral Based on the Definition of $$f(x)$$ Since $$f(x)$$ behaves differently on positive and negative intervals, split the integral at $$x = 0$$: $$-\int_{-5}^3 f(x) \, dx = -\left( \int_{-5}^0 f(x) \, dx + \int_0^3 f(x) \, dx \right)$$ For $$x \in [-5, 0)$$, $$f(x) = -1$$, and for $$x \in (0, 3]$$, $$f(x) = 1$$. ### Step 4: Evaluate Each Integral 1. **Evaluate $$\int_{-5}^0 f(x) \, dx$$:** $$\int_{-5}^0 f(x) \, dx = \int_{-5}^0 (-1) \, dx = -\int_{-5}^0 1 \, dx$$ $$= -(x) \Big|_{-5}^0 = -(0 - (-5)) = -5$$ 2. **Evaluate $$\int_0^3 f(x) \, dx$$:** $$\int_0^3 f(x) \, dx = \int_0^3 1 \, dx = x \Big|_0^3 = 3 - 0 = 3$$ ### Step 5: Combine the Results Now we substitute back: $$-\left( \int_{-5}^0 f(x) \, dx + \int_0^3 f(x) \, dx \right) = -(-5 + 3) = -(-2) = 2$$ ### Final Answer The value of the integral is: $$\boxed{2}$$ The correct answer is therefore (B) $$2$$. --- Would you like further explanation on any of the steps? Here are some related questions you might find useful: 1. How does the function $$f(x)$$ behave for different values of $$x$$? 2. Why is it necessary to split the integral at $$x = 0$$? 3. What is the significance of reversing the limits in an integral? 4. How do you interpret an integral of a piecewise function? 5. How would the result change if the limits were from $$-3$$ to $$5$$? **Tip:** When integrating functions that change behavior over different intervals, splitting the integral at those key points often simplifies the calculation process.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Integration
Piecewise functions
Absolute value
Formulas
Integral of a constant function
Theorems
Definition of definite integrals
Property of integrals with reversed limits
Suitable Grade Level
Grades 11-12
Related Recommendation
Evaluating a Definite Integral for a Piecewise Function from 1 to 5
Definite Integral with Absolute Value: ∫ from 1 to 3 of |5x - 4| dx Solution
Evaluating the Integral \( \int_0^5 |f'(x)||f(x)|^2 \, dx \) from a Graph
Solve Integrals with Even and Odd Functions Symmetry
Calculate the Area Between Absolute Value Functions: f(x) = 5 - |x - 3| and f(x) = |x|