Math Problem Statement
Solve question 40
Solve question 40 from the image provided.
Solution
We are tasked to solve the definite integral in question 40:
Step-by-Step Solution:
1. Handle the Absolute Value Function
The function is defined as:
\begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$$ Thus, $$1 - |x|$$ becomes: $$1 - |x| = \begin{cases} 1 - x, & \text{if } x \geq 0 \\ 1 + x, & \text{if } x < 0 \end{cases}$$ #### 2. Split the Integral Because $$|x|$$ has a different form for $$x \geq 0$$ and $$x < 0$$, split the integral into two parts: $$\int_{-2}^2 (1 - |x|) dx = \int_{-2}^0 (1 + x) dx + \int_{0}^2 (1 - x) dx$$ #### 3. Evaluate Each Part **For $$\int_{-2}^0 (1 + x) dx$$:** $$\int_{-2}^0 (1 + x) dx = \int_{-2}^0 1 \, dx + \int_{-2}^0 x \, dx$$ \[ = \left[ x \right]_{-2}^0 + \left[ \frac{x^2}{2} \right]_{-2}^0 \] $$= \left( 0 - (-2) \right) + \left( \frac{0^2}{2} - \frac{(-2)^2}{2} \right)$$ $$= 2 - 2 = 0$$ **For $$\int_{0}^2 (1 - x) dx$$:** $$\int_{0}^2 (1 - x) dx = \int_{0}^2 1 \, dx - \int_{0}^2 x \, dx$$ \[ = \left[ x \right]_0^2 - \left[ \frac{x^2}{2} \right]_0^2 \] $$= \left( 2 - 0 \right) - \left( \frac{2^2}{2} - \frac{0^2}{2} \right)$$ $$= 2 - 2 = **Let me fully reattemptThe integral we need to solve is: \[ \int_{-2}^2 (1 - |x|) dx$$ ### Step-by-Step Solution #### 1. Handle the Absolute Value The function $$|x|$$ splits the integral at $$x = 0$$, since: $$|x| = \begin{cases} x & \text{if } x \geq 0, \\ -x & \text{if } x < 0. \end{cases}$$ So, $$1 - |x|$$ becomes: $$1 - |x| = \begin{cases} 1 - x & \text{if } x \geq 0, \\ 1 + x & \text{if } x < 0. \end{cases}$$ #### 2. Split the Integral We split the integral into two regions: $$\int_{-2}^2 (1 - |x|) dx = \int_{-2}^0 (1 + x) dx + \int_{0}^2 (1 - x) dx.$$ #### 3. Evaluate Each Part ##### For $$\int_{-2}^0 (1 + x) dx$$: $$\int_{-2}^0 (1 + x) dx = \int_{-2}^0 1 \, dx + \int_{-2}^0 x \, dx.$$ Compute each term: $$\int_{-2}^0 1 \, dx = \left[ x \right]_{-2}^0 = 0 - (-2) = 2,$$ $$\int_{-2}^0 x \, dx = \left[ \frac{x^2}{2} \right]_{-2}^0 = \frac{0^2}{2} - \frac{(-2)^2}{2} = 0 - 2 = -2.$$ So: $$\int_{-2}^0 (1 + x) dx = 2 - 2 = 0.$$ ##### For $$\int_{0}^2 (1 - x) dx$$: $$\int_{0}^2 (1 - x) dx = \int_{0}^2 1 \, dx - \int_{0}^2 x \, dx.$$ Compute each term: $$\int_{0}^2 1 \, dx = \left[ x \right]_0^2 = 2 - 0 = 2,$$ $$\int_{0}^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = 2 - 0 = 2.$$ So: $$\int_{0}^2 (1 - x) dx = 2 - 2 = 0.$$ #### 4. Combine the Results The total integral is: $$\int_{-2}^2 (1 - |x|) dx = \int_{-2}^0 (1 + x) dx + \int_{0}^2 (1 - x) dx = 0 + 0 = 0.$$ ### Final Answer: $$\int_{-2}^2 (1 - |x|) dx = 0$$ --- Would you like additional details or examples? 😊 ### Related Questions: 1. How does symmetry affect definite integrals? 2. What is the geometric interpretation of $$1 - |x|$$ over $$[-2, 2]$$? 3. Can we solve $$\int_{-a}^a (1 - |x|) dx$$ for any $$a > 0$$? 4. How do you compute definite integrals of piecewise functions? 5. How do absolute values modify the bounds of integration? ### Tip: Always check for symmetry in functions when solving definite integrals—it can simplify the calculations significantly!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Definite Integration
Absolute Value Function
Piecewise Functions
Formulas
\( |x| = x \text{ if } x \geq 0; |x| = -x \text{ if } x < 0 \)
Definite Integral Formula: \( \int_a^b f(x) dx \)
Theorems
Properties of Definite Integrals
Handling Absolute Value Functions in Integrals
Suitable Grade Level
Grades 11-12 (Advanced High School Mathematics)