Let's solve the problem **step by step**: ## **Given:** - \( f(x) = x^2 \) on \( -2 \leq x \leq 2 \) - \( f(x) \) is periodic with period \( 4 \) (\( f(x) = f(x+4) \)) - The coefficient \( A_2 \) of the **cosine term** in the Fourier series is being asked. ## **Fourier Coefficient Formula:** For a function of period \(2L\), the Fourier cosine coefficients are: \[ A_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx \] Here, period \(= 4 \implies L = 2\), so: \[ A_n = \frac{1}{2} \int_{-2}^{2} x^2 \cos\left(\frac{n\pi x}{2}\right) dx \] We are asked for \(A_2\): \[ A_2 = \frac{1}{2} \int_{-2}^{2} x^2 \cos\left(\frac{2\pi x}{2}\right) dx = \frac{1}{2} \int_{-2}^{2} x^2 \cos(\pi x) dx \] ## **Evaluate the Integral:** Note: \(x^2 \cos(\pi x)\) is even, so \[ A_2 = \frac{1}{2} \cdot 2 \int_{0}^{2} x^2 \cos(\pi x) dx = \int_{0}^{2} x^2 \cos(\pi x) dx \] Let’s use integration by parts: Let \( u = x^2 \), \( dv = \cos(\pi x) dx \) Then \( du = 2x dx \), \( v = \frac{1}{\pi} \sin(\pi x) \) Apply integration by parts: \[ \int x^2 \cos(\pi x) dx = x^2 \cdot \frac{1}{\pi} \sin(\pi x) - \int \frac{1}{\pi} \sin(\pi x) \cdot 2x dx \] \[ = \frac{x^2}{\pi} \sin(\pi x) - \frac{2}{\pi} \int x \sin(\pi x) dx \] Now, integrate \( \int x \sin(\pi x) dx \) by parts again: Let \( u = x \), \( dv = \sin(\pi x) dx \) So \( du = dx \), \( v = -\frac{1}{\pi} \cos(\pi x) \) \[ \int x \sin(\pi x) dx = -\frac{x}{\pi} \cos(\pi x) + \int \frac{1}{\pi} \cos(\pi x) dx = -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x) \] So, plugging back: \[ \int x^2 \cos(\pi x) dx = \frac{x^2}{\pi} \sin(\pi x) - \frac{2}{\pi} \left( -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x) \right) \] \[ = \frac{x^2}{\pi} \sin(\pi x) + \frac{2x}{\pi^2} \cos(\pi x) - \frac{2}{\pi^3} \sin(\pi x) \] Now, evaluate from \(0\) to \(2\): \[ A_2 = \left[ \frac{x^2}{\pi} \sin(\pi x) + \frac{2x}{\pi^2} \cos(\pi x) - \frac{2}{\pi^3} \sin(\pi x) \right]_{x=0}^{x=2} \] Calculate at \(x=2\): - \(\sin(2\pi) = 0\) - \(\cos(2\pi) = 1\) So at \(x=2\): \[ \frac{(2)^2}{\pi} \cdot 0 + \frac{2 \times 2}{\pi^2} \cdot 1 - \frac{2}{\pi^3} \cdot 0 = \frac{4}{\pi^2} \] At \(x=0\): - \(\sin(0) = 0\) - \(\cos(0) = 1\) So at \(x=0\): \[ 0 + 0 - 0 = 0 \] Thus, \[ A_2 = \frac{4}{\pi^2} - 0 = \boxed{\frac{4}{\pi^2}} \] ## **Final Answer:** \[ \boxed{a) \ \frac{4}{\pi^2}} \]

Let's analyze all the options carefully and understand how they relate to our calculation. --- ### **Given Data Recap:** - Function: \(f(x) = x^2\) for \(-2 < x < 2\) - Period: \(T = 4\), so \(f(x) = f(x+4)\) - Fourier series involves coefficients \(A_n\) for cosine terms: \[ A_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \left( \frac{n \pi x}{L} \right) dx \] with \(L=2\). --- ### **Calculation of \(A_2\):** We computed: \[ A_2 = \frac{1}{2} \int_{-2}^{2} x^2 \cos \left( \frac{2 \pi x}{2} \right) dx = \frac{1}{2} \int_{-2}^{2} x^2 \cos(\pi x) dx \] Since \(x^2\) is even and \(\cos(\pi x)\) is even, the integrand is even, so: \[ A_2 = \int_{0}^{2} x^2 \cos(\pi x) dx \] Using integration by parts, the integral evaluates to: \[ A_2 = \frac{4}{\pi^2} \] --- ### **Now, examining the options:** --- ### **Option a)** \[ A_2 = \frac{4}{\pi^2} \] **This matches our derived value.** --- ### **Option b)** \[ A_2 = \text{Dye} \ 2 \ 17 \] **This seems to be a misprint or nonsensical notation. Possibly an OCR error or typo.** It doesn't match any standard mathematical expression and doesn't relate to our derived value. --- ### **Option c)** \[ A_2 = E \ nm^2 \] **Again, this appears to be meaningless in the context of Fourier coefficients.** It doesn't correspond to our calculation. --- ### **Option d)** \[ A_2 = \text{im}\{ \phi \} \] **"im" likely refers to the imaginary part of some complex expression, but since \(A_2\) is a real coefficient (as the integral is real and produces a real result), this is inconsistent.** --- ### **Option e)** **"None of the other options are correct"** - Since option a) matches our calculation exactly, this statement is false. --- ### **Final conclusion:** - The **correct** value of \(A_2\) is \(\boxed{\frac{4}{\pi^2}}\), which corresponds to **option a)**. - Other options are either nonsensical or irrelevant to the calculation. --- ### **Summary:** | Option | Explanation | Correct? | |----------|--------------------------------------------------------------|----------------------| | a) | \(\frac{4}{\pi^2}\), matches our calculation | **Yes** | | b) | Nonsensical or typo; no relation | No | | c) | Nonsensical; not related to Fourier coefficient | No | | d) | Imaginary part of a complex expression; irrelevant | No | | e) | "None of the other options are correct" | **No**, since a) is correct | --- ### **Final Answer:** \[ \boxed{\textbf{a) } \frac{4}{\pi^2}} \] --- **Hope this detailed explanation clarifies the problem and the reasoning behind the correct option!**