Let's **break down the Fourier series coefficients** based on the given integral and function definition. ### **Given:** \[ \frac{1}{1.5} \left[ \int_{0}^{2} \frac{4e^{x/2}-1}{e^2-1} \sin\left(\frac{n\pi x}{1.5}\right) dx + \int_{2}^{4} 4\sin\left(\frac{n\pi x}{1.5}\right) dx + \int_{4}^{6} (2x+12)\sin\left(\frac{n\pi x}{1.5}\right) dx \right] \] You are asked to find the **period, \(L\), \(a_0\), \(a_n\), and \(b_n\)** for the piecewise function. --- ## **Step 1: Identify the period and \(L\)** From the argument of \(\sin\left(\frac{n\pi x}{1.5}\right)\), we see the period is: \[ \text{Period} = 3 \quad (\text{since } L = 1.5) \] --- ## **Step 2: Write the general Fourier Series** For a function with period \(2L\), the Fourier series is: \[ f(x) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right] \] Here, \(L = 1.5\), so period \(= 3\). --- ## **Step 3: Identify the function \(f(x)\) over \(0 \leq x \leq 6\) (one period)** From the integrals, the piecewise function is: \[ f(x) = \begin{cases} \frac{4e^{x/2}-1}{e^2-1} & 0 \leq x < 2 \\ 4 & 2 \leq x < 4 \\ 2x + 12 & 4 \leq x < 6 \end{cases} \] --- ## **Step 4: Compute coefficients** ### **(a) \(a_0\) coefficient** \[ a_0 = \frac{1}{2L} \int_0^{2L} f(x)\, dx \] Here, \(2L = 3\), but your limits are \(0\) to \(6\), so period is \(6\) and \(L=3\). So, \[ a_0 = \frac{1}{6} \left[ \int_0^2 \frac{4e^{x/2}-1}{e^2-1} dx + \int_2^4 4 dx + \int_4^6 (2x+12) dx \right] \] #### **Evaluate each integral:** 1. \(\int_0^2 \frac{4e^{x/2}-1}{e^2-1} dx\) 2. \(\int_2^4 4 dx = 4 \times (4-2) = 8\) 3. \(\int_4^6 (2x+12) dx\) Let's solve these: #### 1. \(\int_0^2 \frac{4e^{x/2}-1}{e^2-1} dx\): Let \(I_1 = \int_0^2 \frac{4e^{x/2}-1}{e^2-1} dx\) \[ = \frac{1}{e^2-1} \int_0^2 (4e^{x/2} - 1) dx \] \[ = \frac{1}{e^2-1} \left[ 4 \int_0^2 e^{x/2} dx - \int_0^2 dx \right] \] \[ \int e^{x/2} dx = 2e^{x/2} \] So, \[ = \frac{1}{e^2-1} \left[ 4 (2e^{x/2} \Big|_0^2) - (x|_0^2) \right] \] \[ = \frac{1}{e^2-1} \left[ 8(e^{1}-1) - 2 \right] \] #### 2. \(\int_4^6 (2x+12) dx\) \[ = \int_4^6 2x dx + \int_4^6 12 dx \] \[ = 2 \left[ \frac{x^2}{2} \Big|_4^6 \right] + 12(x|_4^6) = (x^2|_4^6) + 12(6-4) = (36-16) + 24 = 20 + 24 = 44 \] --- #### **Now plug these into \(a_0\):** \[ a_0 = \frac{1}{6} \left\{ \frac{8(e-1) - 2}{e^2-1} + 8 + 44 \right\} = \frac{1}{6} \left[ \frac{8(e-1) - 2}{e^2-1} + 52 \right] \] --- ### **(b) \(a_n\) coefficient** \[ a_n = \frac{1}{L} \int_0^{2L} f(x) \cos\left( \frac{n\pi x}{L} \right) dx \] Here, \(L=3\), so argument is \(\frac{n\pi x}{3}\). #### **Given function has only sine terms in the integral, so the Fourier sine coefficients are being calculated (i.e., \(b_n\)), not \(a_n\).** --- ### **(c) \(b_n\) coefficient** \[ b_n = \frac{1}{L} \left[ \int_0^2 \frac{4e^{x/2}-1}{e^2-1} \sin\left( \frac{n\pi x}{3} \right) dx + \int_2^4 4 \sin\left( \frac{n\pi x}{3} \right) dx + \int_4^6 (2x+12) \sin\left( \frac{n\pi x}{3} \right) dx \right] \] Where \(L = 3\). --- ## **Summary Table** | Parameter | Value/Expression | |---|---| | Period | \(6\) | | \(L\) | \(3\) | | \(a_0\) | \(\frac{1}{6} \left[ \frac{8(e-1) - 2}{e^2-1} + 52 \right]\) | | \(a_n\) | \(\frac{1}{3} \left[ \int_0^2 \frac{4e^{x/2}-1}{e^2-1} \cos\left( \frac{n\pi x}{3} \right) dx + \int_2^4 4 \cos\left( \frac{n\pi x}{3} \right) dx + \int_4^6 (2x+12) \cos\left( \frac{n\pi x}{3} \right) dx \right]\) | | \(b_n\) | \(\frac{1}{3} \left[ \int_0^2 \frac{4e^{x/2}-1}{e^2-1} \sin\left( \frac{n\pi x}{3} \right) dx + \int_2^4 4 \sin\left( \frac{n\pi x}{3} \right) dx + \int_4^6 (2x+12) \sin\left( \frac{n\pi x}{3} \right) dx \right]\) | --- ## **Final Answers (in summary):** - **Period:** 6 - **L:** 3 - **\(a_0\):** \(\frac{1}{6} \left[ \frac{8(e-1) - 2}{e^2-1} + 52 \right]\) - **\(a_n\):** (see table above) - **\(b_n\):** (see table above) If you want the explicit evaluation of any integral, let me know!