91Ó°ÊÓ

First, let's find the coefficient a_0 using the given formula. Since f(x) = x, we have: \(a_0 = \frac{2}{c} \int_{0}^{c} x dx\) Integrating x with respect to x, we get: \(a_0 = \frac{2}{c} \left[\frac{x^2}{2} \right]_0^{c}\) Evaluating the integral and simplifying, we obtain: \(a_0 = \frac{2}{c} \cdot \frac{c^2}{2} - \frac{2}{c} \cdot \frac{0^2}{2} = c\) So, the constant term a_0 in our Fourier cosine series is equal to c.

Next, let's calculate the coefficients a_n using the given formula. Following the same process as for a_0, we have: \(a_n = \frac{2}{c} \int_{0}^{c} x \cos{\frac{n \pi x}{c}} dx\) To evaluate this integral, we'll use integration by parts: Let u = x, and dv = \(\cos{\frac{n \pi x}{c}} dx\). Then du = dx, and v = \(\frac{c}{n \pi} \sin{\frac{n \pi x}{c}}\). Now, applying the integration by parts formula: \(a_n = \frac{2}{c} \left[u \cdot v - \int v du \right]_0^{c}\) Substitute the expressions for u, v, and du: \(a_n = \frac{2}{c} \left[x \cdot \frac{c}{n \pi} \sin{\frac{n \pi x}{c}} - \int \frac{c}{n \pi} \sin{\frac{n \pi x}{c}} dx \right]_0^{c}\) Simplifying the integral, we get: \(a_n = \frac{2}{n \pi} \left[c \cdot x \sin{\frac{n \pi x}{c}} - \int c \sin{\frac{n \pi x}{c}} dx \right]_0^{c}\) Integrating \(\sin{\frac{n \pi x}{c}}\) with respect to x, we have: \(a_n = \frac{2}{n \pi} \left[c \cdot x \sin{\frac{n \pi x}{c}} - \frac{-c^2}{n \pi} \cos{\frac{n \pi x}{c}} \right]_0^{c}\) Evaluating the resulting expression and simplifying, we get: \(a_n = \frac{2}{n \pi} \left( \frac{c^2}{n \pi} \right) = \frac{2c^2}{n^2 \pi^2}\)

Finally, we write the Fourier cosine series using the coefficients we found in the previous steps: \(f(x) \approx \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos{\frac{n \pi x}{c}}\) Substituting our values for a_0 and a_n, we obtain: \(f(x) \approx \frac{c}{2} + \sum_{n=1}^{\infty} \frac{2c^2}{n^2 \pi^2} \cos{\frac{n \pi x}{c}}\) This is the Fourier cosine series representation of the function f(x) = x over the interval 0 < x < c.