Math Problem Statement
Evaluate the integral ∫ from -2 to 1 of |x| dx
Solution
The given integral is:
To solve this, we need to handle the absolute value function. The absolute value function, , behaves differently depending on whether is positive or negative. Specifically:
\begin{cases} -x & \text{if } x < 0 \\ x & \text{if } x \geq 0 \end{cases}$$ So, we will break the integral into two parts: 1. From $$-2$$ to $$0$$, where $$|x| = -x$$. 2. From $$0$$ to $$1$$, where $$|x| = x$$. The integral becomes: $$\int_{-2}^{1} |x| \, dx = \int_{-2}^{0} -x \, dx + \int_{0}^{1} x \, dx$$ Now, let's compute these integrals separately. 1. $$\int_{-2}^{0} -x \, dx$$ $$= \left[ -\frac{x^2}{2} \right]_{-2}^{0} = -\frac{(0)^2}{2} + \frac{(-2)^2}{2} = 0 + \frac{4}{2} = 2$$ 2. $$\int_{0}^{1} x \, dx$$ $$= \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{(1)^2}{2} - \frac{(0)^2}{2} = \frac{1}{2} - 0 = \frac{1}{2}$$ Now, adding these results together: $$2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$$ So, the value of the integral is: $$\int_{-2}^{1} |x| \, dx = \frac{5}{2}$$ Would you like any further details or clarification? Here are some related questions you could explore: 1. How does the absolute value function affect integration? 2. How do we handle improper integrals involving absolute values? 3. What is the geometric interpretation of this integral? 4. How do you compute integrals involving piecewise functions? 5. What happens if the limits of integration include singular points? **Tip:** When integrating functions with absolute values, breaking the integral into intervals where the expression is continuous is key!Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Integration
Absolute Value Function
Piecewise Functions
Formulas
∫ |x| dx = ∫ -x dx (for x < 0) + ∫ x dx (for x ≥ 0)
Theorems
Fundamental Theorem of Calculus
Properties of Definite Integrals
Suitable Grade Level
Grades 11-12 or Early College