Math Problem Statement
Solution
The given integral is:
To evaluate this integral, we need to handle the absolute value function , which behaves differently for and . Here are the steps:
Step 1: Split the integral based on the behavior of
The absolute value of is defined as:
\begin{cases} x, & \text{if } x \geq 0, \\ -x, & \text{if } x < 0. \end{cases}$$ The range of integration, $$[-1, 3]$$, crosses $$x = 0$$. Therefore, we split the integral into two parts: $$\int_{-1}^{3} (|x| - 1) \, dx = \int_{-1}^{0} ((-x) - 1) \, dx + \int_{0}^{3} (x - 1) \, dx.$$ --- ### Step 2: Evaluate each part 1. **Evaluate $$\int_{-1}^{0} ((-x) - 1) \, dx$$:** Simplify the integrand: $$\int_{-1}^{0} ((-x) - 1) \, dx = \int_{-1}^{0} (-x - 1) \, dx.$$ Compute the integral: $$\int_{-1}^{0} (-x - 1) \, dx = \left[-\frac{x^2}{2} - x\right]_{-1}^{0}.$$ Substituting the limits: $$= \left(-\frac{0^2}{2} - 0\right) - \left(-\frac{(-1)^2}{2} - (-1)\right) = 0 - \left(-\frac{1}{2} + 1\right) = 0 - \frac{1}{2} + 1 = \frac{1}{2}.$$ 2. **Evaluate $$\int_{0}^{3} (x - 1) \, dx$$:** Simplify the integrand: $$\int_{0}^{3} (x - 1) \, dx = \int_{0}^{3} x \, dx - \int_{0}^{3} 1 \, dx.$$ Compute the integral: $$\int_{0}^{3} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{3} = \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2},$$ and $$\int_{0}^{3} 1 \, dx = \left[x\right]_{0}^{3} = 3 - 0 = 3.$$ Combine the results: $$\int_{0}^{3} (x - 1) \, dx = \frac{9}{2} - 3 = \frac{3}{2}.$$ --- ### Step 3: Add the results Add the two parts together: $$\int_{-1}^{3} (|x| - 1) \, dx = \frac{1}{2} + \frac{3}{2} = 2.$$ --- ### Final Answer: $$\boxed{2}$$ Would you like a deeper explanation of any step? --- ### Related Questions: 1. How does the behavior of the absolute value function affect integral evaluation? 2. Can we solve this integral geometrically instead of algebraically? 3. What if the limits of integration were different, such as $$[-3, 3]$$? 4. How does the choice of splitting points influence the computation process? 5. Can we verify the result using numerical integration techniques? ### Tip: When working with absolute values in integrals, always split the integral at the points where the argument of the absolute value function changes sign. This ensures accurate computation.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Integral Calculus
Absolute Value Functions
Formulas
Definition of the definite integral: \( \int_a^b f(x) \, dx \)
Absolute value function: \( |x| = x \text{ if } x \geq 0, \text{ and } |x| = -x \text{ if } x < 0 \)
Theorems
Properties of definite integrals
Suitable Grade Level
Grades 11-12 or early college
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