Define the functions 𝑓 , 𝑔: R → R as 𝑓 (𝑥) := 1 if0 <𝑥 < 1, 0 else,    and 𝑔(𝑥) :=      𝑥 if 0 < 𝑥 ≤ 1, 2 −𝑥 if1<𝑥 <2, 0 else. Compute the Fourier transform of 𝑔.

To compute the Fourier transform of the function \( g(x) \), we will proceed by analyzing \( g(x) \) as defined in different intervals. The function \( g(x) \) is defined as:

\[
g(x) = 
\begin{cases} 
x & \text{if } 0 < x \leq 1, \\
2 - x & \text{if } 1 < x < 2, \\
0 & \text{elsewhere}.
\end{cases}
\]

### Fourier Transform Definition

The Fourier transform of \( g(x) \), denoted \( \hat{g}(k) \), is given by:
\[
\hat{g}(k) = \int_{-\infty}^{\infty} g(x) e^{-2 \pi i k x} \, dx.
\]

Since \( g(x) \) is zero outside \( (0, 2) \), we can restrict the limits of integration:
\[
\hat{g}(k) = \int_0^1 x e^{-2 \pi i k x} \, dx + \int_1^2 (2 - x) e^{-2 \pi i k x} \, dx.
\]

### Step 1: Compute the First Integral

We compute:
\[
\int_0^1 x e^{-2 \pi i k x} \, dx.
\]
Using integration by parts, let:
- \( u = x \Rightarrow du = dx \),
- \( dv = e^{-2 \pi i k x} dx \Rightarrow v = \frac{e^{-2 \pi i k x}}{-2 \pi i k} \).

Then,
\[
\int_0^1 x e^{-2 \pi i k x} \, dx = \left. \frac{-x e^{-2 \pi i k x}}{2 \pi i k} \right|_0^1 + \int_0^1 \frac{e^{-2 \pi i k x}}{2 \pi i k} \, dx.
\]
Evaluating at \( x = 1 \) and \( x = 0 \), we get:
\[
= \frac{-e^{-2 \pi i k}}{2 \pi i k} + \frac{1}{(2 \pi i k)^2} \left( e^{-2 \pi i k x} \right)_0^1.
\]
Substitute and simplify this result.

### Step 2: Compute the Second Integral

Now, we compute:
\[
\int_1^2 (2 - x) e^{-2 \pi i k x} \, dx.
\]


Define the functions f, g: R → R as f(x) := 1 if 0 < x < 1, 0 else, and g(x) := x if 0 < x ≤ 1, 2 − x if 1 < x < 2, 0 else. Compute the Fourier transform of g.

Explore the Fourier transform of a piecewise function g(x) defined by g(x) = x for 0 < x ≤ 1, g(x) = 2 - x for 1 < x < 2, and g(x) = 0 elsewhere. This solution covers integral calculus techniques and Fourier transform concepts, ideal for university-level math students.

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