Math Problem Statement
Definite integrals of piecewise functions Google Classroom Microsoft Teams Problem
[f(x) = \begin{cases} -6x+6 & \text{for} ~~~~x\lt1 \ \sin(\pi x) & \text{for} ~~~~ x \geq1\end{cases}] Evaluate the definite integral. [\displaystyle\int^3_{0}f(x),dx = ] Choose 1 answer: Choose 1 answer: (Choice A)
[-3+\dfrac{2}\pi] A
[-3+\dfrac{2}\pi] (Choice B)
[6-\dfrac{2}\pi] B
[6-\dfrac{2}\pi]
(Choice C)
[3]
C
[3]
(Choice D)
[\dfrac{2}\pi] D
[\dfrac{2}\pi]
Solution
To evaluate the definite integral for the given piecewise function:
\begin{cases} -6x + 6 & \text{for } x < 1 \\ \sin(\pi x) & \text{for } x \geq 1 \end{cases}$$ ### Step 1: Break the integral into two parts We divide the integral at $$x = 1$$ since $$f(x)$$ changes its definition there: $$\int_{0}^{3} f(x) \, dx = \int_{0}^{1} (-6x + 6) \, dx + \int_{1}^{3} \sin(\pi x) \, dx.$$ ### Step 2: Solve $$\int_{0}^{1} (-6x + 6) \, dx$$ The antiderivative of $$-6x + 6$$ is: $$\int (-6x + 6) \, dx = -3x^2 + 6x + C.$$ Evaluating from $$0$$ to $$1$$: $$\int_{0}^{1} (-6x + 6) \, dx = \left[ -3x^2 + 6x \right]_{0}^{1}.$$ At $$x = 1$$: $$-3(1)^2 + 6(1) = -3 + 6 = 3.$$ At $$x = 0$$: $$-3(0)^2 + 6(0) = 0.$$ Thus: $$\int_{0}^{1} (-6x + 6) \, dx = 3 - 0 = 3.$$ ### Step 3: Solve $$\int_{1}^{3} \sin(\pi x) \, dx$$ The antiderivative of $$\sin(\pi x)$$ is: $$\int \sin(\pi x) \, dx = -\frac{1}{\pi} \cos(\pi x) + C.$$ Evaluating from $$1$$ to $$3$$: $$\int_{1}^{3} \sin(\pi x) \, dx = \left[ -\frac{1}{\pi} \cos(\pi x) \right]_{1}^{3}.$$ At $$x = 3$$: $$-\frac{1}{\pi} \cos(\pi \cdot 3) = -\frac{1}{\pi} \cos(3\pi).$$ Since $$\cos(3\pi) = -1$$, this becomes: $$-\frac{1}{\pi}(-1) = \frac{1}{\pi}.$$ At $$x = 1$$: $$-\frac{1}{\pi} \cos(\pi \cdot 1) = -\frac{1}{\pi} \cos(\pi).$$ Since $$\cos(\pi) = -1$$, this becomes: $$-\frac{1}{\pi}(-1) = \frac{1}{\pi}.$$ Thus: $$\int_{1}^{3} \sin(\pi x) \, dx = \frac{1}{\pi} - \frac{1}{\pi} = 0.$$ ### Step 4: Combine the results The total integral is: $$\int_{0}^{3} f(x) \, dx = \int_{0}^{1} (-6x + 6) \, dx + \int_{1}^{3} \sin(\pi x) \, dx = 3 + 0 = 3.$$ ### Final Answer: $$\boxed{\text{Choice C: } 3}$$ --- Do you have further questions or need clarification? Here are five related questions to expand your understanding: 1. How would the solution change if $$f(x)$$ was defined differently for $$x < 1$$? 2. Can you evaluate the integral if the piecewise function has three intervals instead of two? 3. How does the periodicity of $$\sin(\pi x)$$ affect its integral over various intervals? 4. What would happen if the bounds of the integral were negative, such as $$[-1, 3]$$? 5. How does the Fundamental Theorem of Calculus help in evaluating definite integrals? **Tip:** For piecewise functions, always split the integral at points where the function's definition changes.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Definite Integrals
Piecewise Functions
Trigonometric Functions
Antiderivatives
Formulas
∫ (ax + b) dx = (a/2)x^2 + bx + C
∫ sin(πx) dx = -1/π cos(πx) + C
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
Grades 11-12
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