Math Problem Statement
Find the area of the region enclosed by f(x) and the x-axis for the given function over the specified interval.
f left parenthesis x right parenthesis equals left brace Start 2 By 2 Matrix 1st Row 1st Column x squared plus 2 x plus 2 2nd Column x less than 2 2nd Row 1st Column 3 x plus 4 2nd Column x greater than or equals 2 EndMatrix
on left bracket negative 3 comma 3 right bracket
Solution
To find the area of the region enclosed by the function and the -axis over the interval , we will compute the definite integrals for each piece of the function and sum their absolute values. Here's the step-by-step approach:
1. Function Analysis
The function is piecewise-defined as:
\begin{cases} x^2 + 2x + 2, & x < 2, \\ 3x + 4, & x \geq 2. \end{cases}$$ The given interval is $$[-3, 3]$$, which spans across both pieces of the function. ### 2. Break the Interval - For $$x < 2$$, the function is $$x^2 + 2x + 2$$. - Interval: $$[-3, 2)$$ - For $$x \geq 2$$, the function is $$3x + 4$$. - Interval: $$[2, 3]$$ ### 3. Compute the Definite Integrals The total area is the sum of the absolute values of the integrals: $$\text{Total Area} = \left| \int_{-3}^2 \left( x^2 + 2x + 2 \right) \, dx \right| + \left| \int_{2}^3 \left( 3x + 4 \right) \, dx \right|.$$ #### a. Integral over $$[-3, 2)$$ for $$x^2 + 2x + 2$$: $$\int (x^2 + 2x + 2) \, dx = \frac{x^3}{3} + x^2 + 2x + C$$ Evaluate from $$-3$$ to $$2$$: \[ \int_{-3}^2 (x^2 + 2x + 2) \, dx = \left[ \frac{2^3}{3} + 2^2 + 2(2) \right] - \left[ \frac{(-3)^3}{3} + (-3)^2 + 2(-3) \right] \] \[ = \left[ \frac{8}{3} + 4 + 4 \right] - \left[ \frac{-27}{3} + 9 - 6 \right] \] \[ = \left[ \frac{8}{3} + 8 \right] - \left[ -9 + 9 - 6 \right] \] $$= \frac{8}{3} + 8 - (-6) = \frac{8}{3} + 14 = \frac{50}{3}.$$ #### b. Integral over $$[2, 3]$$ for $$3x + 4$$: $$\int (3x + 4) \, dx = \frac{3x^2}{2} + 4x + C$$ Evaluate from $$2$$ to $$3$$: \[ \int_{2}^3 (3x + 4) \, dx = \left[ \frac{3(3)^2}{2} + 4(3) \right] - \left[ \frac{3(2)^2}{2} + 4(2) \right] \] \[ = \left[ \frac{27}{2} + 12 \right] - \left[ \frac{12}{2} + 8 \right] \] \[ = \left[ \frac{27}{2} + \frac{24}{2} \right] - \left[ \frac{12}{2} + \frac{16}{2} \right] \] $$= \frac{51}{2} - \frac{28}{2} = \frac{23}{2}.$$ ### 4. Total Area Add the absolute values of the areas: $$\text{Total Area} = \left| \frac{50}{3} \right| + \left| \frac{23}{2} \right|$$ $$= \frac{50}{3} + \frac{23}{2}.$$ ### 5. Simplify Find a common denominator: $$\text{Total Area} = \frac{100}{6} + \frac{69}{6} = \frac{169}{6}.$$ ### Final Answer: The area of the region enclosed by $$f(x)$$ and the $$x$$-axis is: $$\boxed{\frac{169}{6}} \text{ square units.}$$ Would you like further details on any step or additional examples? Here are 5 questions to consider next: 1. How does the piecewise nature of a function affect the integration process? 2. What are common strategies for integrating polynomials? 3. How can areas below the $$x$$-axis affect total area calculations? 4. Can you compute the area for a different interval? 5. How do discontinuities in a piecewise function influence integration? **Tip**: Always ensure to compute absolute values when finding the total area enclosed by a curve and the $$x$$-axis.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Integration
Definite Integrals
Piecewise Functions
Absolute Value of Areas
Formulas
Definite Integral: ∫[a, b] f(x) dx
Integral of a Polynomial: ∫(ax^n) dx = a * (x^(n+1) / (n+1)) + C
Absolute Value: |x|
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
Undergraduate or Advanced High School (Grades 11-12)
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