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).

The given function \( f(x) \) is defined piecewise as:

\[
f(x) = 
\begin{cases} 
x & \text{for } x \in [0, 2) \\
5 & \text{for } x = 2 \\
2 & \text{for } x \in (2, 4]
\end{cases}
\]

### (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.

Let f : [0, 4] → R be f(x) = { x for x ∈ [0, 2), 5 for x = 2, 2 for x ∈ (2, 4]. (a) Find a formula for F(x) = ∫_1^x f(t) dt. (b) What is F'(x)? Hint: it is not f(x).

Explore how to find the integral F(x) = ∫_1^x f(t) dt and its derivative F'(x) for a piecewise function defined on [0, 4]. This problem covers key calculus concepts such as integration, differentiation, and the Fundamental Theorem of Calculus, with a focus on handling discontinuities.

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