Math Problem Statement
evaluate the integral -2 to 2 if f(x) = 2 for all x between -2 and 0 and f(x) equals 4-x^2 for all x between 0 and 2
Solution
To evaluate the integral from to for the piecewise function , we'll break the integral into two parts based on the definition of . The function is given as:
\begin{cases} 2 & \text{for } -2 \leq x \leq 0 \\ 4 - x^2 & \text{for } 0 \leq x \leq 2 \end{cases}$$ The total integral can be written as the sum of two integrals: $$\int_{-2}^{2} f(x) \, dx = \int_{-2}^{0} 2 \, dx + \int_{0}^{2} (4 - x^2) \, dx$$ ### Step 1: Evaluate $$\int_{-2}^{0} 2 \, dx$$ Since $$f(x) = 2$$ over the interval $$[-2, 0]$$, this is a simple integral of a constant function: $$\int_{-2}^{0} 2 \, dx = 2[x]_{-2}^{0} = 2(0 - (-2)) = 2 \times 2 = 4$$ ### Step 2: Evaluate $$\int_{0}^{2} (4 - x^2) \, dx$$ Now, we'll compute the integral of $$4 - x^2$$ over $$[0, 2]$$: $$\int_{0}^{2} (4 - x^2) \, dx = \int_{0}^{2} 4 \, dx - \int_{0}^{2} x^2 \, dx$$ The first part: $$\int_{0}^{2} 4 \, dx = 4[x]_{0}^{2} = 4(2 - 0) = 8$$ The second part: $$\int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}$$ So: $$\int_{0}^{2} (4 - x^2) \, dx = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$$ ### Step 3: Add the results Now, sum the two integrals: $$\int_{-2}^{2} f(x) \, dx = 4 + \frac{16}{3} = \frac{12}{3} + \frac{16}{3} = \frac{28}{3}$$ ### Final Answer: $$\boxed{\frac{28}{3}}$$ Let me know if you need further clarification or details! Here are 5 related questions to deepen your understanding: 1. How would the answer change if the limits of integration were from $$-3$$ to $$3$$? 2. Can you express the definite integral graphically using areas? 3. What happens if the function $$f(x)$$ were defined as $$f(x) = x^2$$ for all $$x$$? 4. How do you handle piecewise functions in integrals if they have more than two pieces? 5. Can you apply this method if the upper bound is a variable, say $$x$$, instead of 2? **Tip:** When dealing with piecewise functions, always break the integral into parts corresponding to the intervals where the function is defined separately.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Integral Calculus
Piecewise Functions
Formulas
Integral of a constant function
Integral of a polynomial
Theorems
-
Suitable Grade Level
College Level
Related Recommendation
Calculate the Definite Integral of a Piecewise Function on [0,3]
Evaluate the Definite Integral of a Piecewise Function f(x) from 0 to 8
Area Enclosed by Piecewise Function and X-axis: Step-by-Step Solution
Piecewise Function Analysis with Integrals and Derivatives
Solving Definite Integrals for a Piecewise Function