Math Problem Statement
Let f : [0, 4] → R be f(x) = x x ∈ [0, 2) 5 x = 2 2 x ∈ (2, 4] (a) Find a formula for F(x) = R x 1 f(t) dt. (b) What is F ′ (x)? Hint it is not f(x).
Solution
The given function is defined piecewise as:
\begin{cases} x & \text{for } x \in [0, 2) \\ 5 & \text{for } x = 2 \\ 2 & \text{for } x \in (2, 4] \end{cases}$$ ### (a) Find a formula for $$F(x) = \int_1^x f(t) \, dt$$ To compute $$F(x) = \int_1^x f(t) \, dt$$, we need to consider different cases based on the intervals in which $$x$$ falls because $$f(t)$$ is piecewise. **Case 1: $$x \in [1, 2)$$** In this range, $$f(t) = t$$. So, the integral is: $$F(x) = \int_1^x t \, dt = \left[\frac{t^2}{2}\right]_1^x = \frac{x^2}{2} - \frac{1^2}{2} = \frac{x^2}{2} - \frac{1}{2}$$ Thus, for $$x \in [1, 2)$$: $$F(x) = \frac{x^2}{2} - \frac{1}{2}$$ **Case 2: $$x \in [2, 4]$$** Here, we split the integral at 2 because of the change in the function definition. For $$t \in [1, 2)$$, we know $$f(t) = t$$, and for $$t = 2$$, $$f(t) = 5$$, and for $$t \in (2, x]$$, $$f(t) = 2$$. The integral becomes: $$F(x) = \int_1^2 t \, dt + \int_2^x 2 \, dt$$ We already computed $$\int_1^2 t \, dt$$: $$\int_1^2 t \, dt = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$$ Now for $$\int_2^x 2 \, dt$$: $$\int_2^x 2 \, dt = 2(x - 2)$$ Thus, for $$x \in [2, 4]$$: $$F(x) = \frac{3}{2} + 2(x - 2) = \frac{3}{2} + 2x - 4 = 2x - \frac{5}{2}$$ ### (b) What is $$F'(x)$$? We can differentiate $$F(x)$$ in each of the cases to find $$F'(x)$$. **Case 1: $$x \in [1, 2)$$** $$F(x) = \frac{x^2}{2} - \frac{1}{2}$$ Taking the derivative: $$F'(x) = \frac{d}{dx}\left( \frac{x^2}{2} - \frac{1}{2} \right) = x$$ So, for $$x \in [1, 2)$$, $$F'(x) = x$$, which matches $$f(x)$$ in this interval. **Case 2: $$x \in [2, 4]$$** $$F(x) = 2x - \frac{5}{2}$$ Taking the derivative: $$F'(x) = 2$$ Thus, for $$x \in [2, 4]$$, $$F'(x) = 2$$, which matches $$f(x)$$ in this interval. ### Important Note: At $$x = 2$$, $$f(x) = 5$$, but the derivative $$F'(x)$$ is not equal to 5. The derivative $$F'(x)$$ represents the continuous part of the function, and so at $$x = 2$$, $$F'(x)$$ will not match $$f(x)$$. Hence, $$F'(x)$$ is not simply equal to $$f(x)$$ everywhere. --- Would you like more details on any step? Here are five related questions: 1. How does the Fundamental Theorem of Calculus apply to this piecewise function? 2. Can $$F'(x)$$ ever be equal to $$f(x)$$ at discontinuities? 3. How would $$F(x)$$ change if the piece at $$x = 2$$ were different? 4. How can we handle discontinuities in the context of integrals and derivatives? 5. What is the interpretation of $$F(x)$$ in terms of area under the curve of $$f(x)$$? **Tip:** When dealing with piecewise functions, always ensure you handle discontinuities properly when computing integrals or derivatives.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Piecewise Functions
Integration
Differentiation
Formulas
F(x) = ∫_1^x f(t) dt
Derivative of F(x)
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
Grades 11-12
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