Let's solve **Question 2** step by step. --- ### Question 2 (25 marks) **Approximate \( \int_1^4 \ln x \, dx \) using the trapezoidal rule with 4 strips, to 3 decimal places.** --- #### **Step 1: Identify interval and number of strips** - Lower limit (\(a\)) = 1 - Upper limit (\(b\)) = 4 - Number of strips (\(n\)) = 4 #### **Step 2: Calculate the width of each strip** \[ h = \frac{b-a}{n} = \frac{4-1}{4} = \frac{3}{4} = 0.75 \] #### **Step 3: Find the x-values** \[ x_0 = 1 \\ x_1 = 1 + 0.75 = 1.75 \\ x_2 = 1 + 2(0.75) = 2.5 \\ x_3 = 1 + 3(0.75) = 3.25 \\ x_4 = 4 \] #### **Step 4: Calculate the function values** \[ f(x) = \ln x \] \[ f(x_0) = \ln(1) = 0 \\ f(x_1) = \ln(1.75) \approx 0.5606 \\ f(x_2) = \ln(2.5) \approx 0.9163 \\ f(x_3) = \ln(3.25) \approx 1.1787 \\ f(x_4) = \ln(4) \approx 1.3863 \\ \] #### **Step 5: Apply the Trapezoidal Rule** \[ \int_a^b f(x)\,dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plug in the values: \[ \frac{0.75}{2} \left[ 0 + 2(0.5606) + 2(0.9163) + 2(1.1787) + 1.3863 \right] \] \[ = 0.375 \left[ 0 + 1.1212 + 1.8326 + 2.3574 + 1.3863 \right] \] Sum inside the bracket: \[ 1.1212 + 1.8326 = 2.9538 \\ 2.9538 + 2.3574 = 5.3112 \\ 5.3112 + 1.3863 = 6.6975 \] So, \[ \text{Approximate value} = 0.375 \times 6.6975 = 2.5116 \] #### **Step 6: Round to 3 decimal places** \[ \boxed{2.512} \] --- ## **Final Answer (to 3 decimal places):** \[ \int_1^4 \ln x \, dx \approx \boxed{2.512} \]

Certainly! Here's the detailed step-by-step solution for **Question 2** as previously provided, along with the questions **Question 4** parts (a), (b), and (c). --- ### **Question 2 (25 marks)** **Approximate \( \int_1^4 \ln x \, dx \) using the trapezoidal rule with 4 strips, to 3 decimal places.** --- ### **Solution:** **Step 1: Identify the interval and number of strips** - \(a = 1\) - \(b = 4\) - Number of strips, \(n = 4\) **Step 2: Calculate the width of each strip** \[ h = \frac{b - a}{n} = \frac{4 - 1}{4} = \frac{3}{4} = 0.75 \] **Step 3: Find the x-values** \[ x_0 = 1 \\ x_1 = 1 + 0.75 = 1.75 \\ x_2 = 1 + 2 \times 0.75 = 2.5 \\ x_3 = 1 + 3 \times 0.75 = 3.25 \\ x_4 = 4 \] **Step 4: Calculate the function values at these points** \[ f(x) = \ln x \] \[ f(x_0) = \ln(1) = 0 \\ f(x_1) = \ln(1.75) \approx 0.5606 \\ f(x_2) = \ln(2.5) \approx 0.9163 \\ f(x_3) = \ln(3.25) \approx 1.1787 \\ f(x_4) = \ln(4) \approx 1.3863 \] **Step 5: Apply the trapezoidal rule formula** \[ \int_a^b f(x)\, dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plug in the values: \[ = \frac{0.75}{2} \left[ 0 + 2(0.5606) + 2(0.9163) + 2(1.1787) + 1.3863 \right] \] Calculate inside the brackets: \[ = 0.375 \times \left[ 0 + 1.1212 + 1.8326 + 2.3574 + 1.3863 \right] \] Sum: \[ 1.1212 + 1.8326 = 2.9538 \\ 2.9538 + 2.3574 = 5.3112 \\ 5.3112 + 1.3863 = 6.6975 \] Calculate the final approximation: \[ 0.375 \times 6.6975 = 2.5116 \] **Step 6: Round to three decimal places** \[ \boxed{2.512} \] --- ### **Final Answer:** \[ \boxed{\int_1^4 \ln x \, dx \approx \mathbf{2.512}} \] --- ### **Question 4** a) **Find the Taylor series of the function \(f(x) = \ln x\) centered at \(x_0 = 1\).** b) **Find the Maclaurin series for the function \(f(x) = \sin x\).** c) **Determine for which values of \(x\) the series \(\sum_{n=0}^\infty \frac{(2x)^n}{n!}\) converges, specifically at \(x = 1\).** --- ### **Answer:** **a) Taylor series of \(f(x) = \ln x\) centered at \(x_0 = 1\):** The derivatives: \[ f(x) = \ln x \\ f'(x) = \frac{1}{x} \\ f''(x) = -\frac{1}{x^2} \\ f'''(x) = \frac{2}{x^3} \\ \text{and so on.} \] At \(x=1\): \[ f(1) = 0 \\ f'(1) = 1 \\ f''(1) = -1 \\ f'''(1) = 2 \\ f^{(4)}(1) = -6 \\ \] Pattern: \[ f^{(n)}(1) = (-1)^{n-1} (n-1)! \] Thus, the Taylor series about \(x=1\): \[ \ln x = \sum_{n=1}^\infty \frac{f^{(n)}(1)}{n!} (x-1)^n = \sum_{n=1}^\infty \frac{(-1)^{n-1} (n-1)!}{n!} (x-1)^n \] Since \(\frac{(n-1)!}{n!} = \frac{1}{n}\), \[ \boxed{ \ln x = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n } \] valid for \(|x-1| < 1\). --- **b) Maclaurin series for \(f(x) = \sin x\):** The derivatives: \[ f(x) = \sin x \\ f'(x) = \cos x \\ f''(x) = -\sin x \\ f'''(x) = -\cos x \\ f^{(4)}(x) = \sin x \] At \(x=0\): \[ \sin 0=0, \quad \cos 0=1 \] Series: \[ \sin x = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n \] The pattern: - For even \(n\), derivatives are 0 at 0. - For odd \(n=2k+1\), derivatives alternate between 1 and -1. Thus, \[ \boxed{ \sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} } \] --- **c) Convergence of the series \(\sum_{n=0}^\infty \frac{(2x)^n}{n!}\):** This is the expansion of \(e^{2x}\), which converges for all real \(x\) (entire function). At \(x=1\): \[ \sum_{n=0}^\infty \frac{(2 \times 1)^n}{n!} = \sum_{n=0}^\infty \frac{2^n}{n!} = e^{2} \] which converges. **Therefore, the series converges for all real \(x\), including at \(x=1\).** --- ### **Summary:** | Part | Description | Series/Result | Valid for | |-------|--------------|----------------|------------| | (a) | Taylor series of \(\ln x\) at \(x=1\) | \(\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(x-1)^n\) | \(|x-1| < 1\) | | (b) | Maclaurin series of \(\sin x\) | \(\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\) | All \(x\) | | (c) | Convergence of \(\sum_{n=0}^\infty \frac{(2x)^n}{n!}\) | Converges for all real \(x\), including \(x=1\) | All real \(x\) | --- Let me know if you'd like further explanations!

Certainly! Here's a detailed step-by-step solution for **Question 3**: --- ### Question 3 (25 marks) **Find the root of the equation \(f(x) = x^3 - x - 1\) using the Bisection method. Given the interval \([1, 2]\). Perform six (6) iterations. Write your answers to 4 decimal places.** --- ### Step 1: Understand the problem We need to find a root of \(f(x) = x^3 - x - 1\) within the interval \([1, 2]\), using the Bisection method. --- ### Step 2: Verify the initial interval Calculate \(f(1)\) and \(f(2)\): \[ f(1) = 1^3 - 1 - 1 = -1 \\ f(2) = 8 - 2 - 1 = 5 \] Since \(f(1) < 0\) and \(f(2) > 0\), a root exists between 1 and 2. --- ### Step 3: Bisection method process At each iteration: - Compute \(x_{mid} = \frac{x_{low} + x_{high}}{2}\) - Evaluate \(f(x_{mid})\) - Decide whether the root lies in \([x_{low}, x_{mid}]\) or \([x_{mid}, x_{high}]\) --- ### **Iteration 1** - \(x_{low} = 1\), \(x_{high} = 2\) \[ x_{mid} = \frac{1 + 2}{2} = 1.5 \] Calculate \(f(1.5)\): \[ f(1.5) = (1.5)^3 - 1.5 - 1 = 3.375 - 1.5 - 1 = 0.875 \] Since \(f(1.5) > 0\), the root is between 1 and 1.5. Update: - \(x_{low} = 1\) - \(x_{high} = 1.5\) --- ### **Iteration 2** \[ x_{mid} = \frac{1 + 1.5}{2} = 1.25 \] Calculate \(f(1.25)\): \[ f(1.25) = (1.25)^3 - 1.25 - 1 = 1.953125 - 1.25 - 1 = -0.296875 \] Since \(f(1.25) < 0\), the root is between 1.25 and 1.5. Update: - \(x_{low} = 1.25\) - \(x_{high} = 1.5\) --- ### **Iteration 3** \[ x_{mid} = \frac{1.25 + 1.5}{2} = 1.375 \] Calculate \(f(1.375)\): \[ f(1.375) = (1.375)^3 - 1.375 - 1 \\ = 2.5996 - 1.375 - 1 = 0.2246 \] Since \(f(1.375) > 0\), root between 1.25 and 1.375. Update: - \(x_{low} = 1.25\) - \(x_{high} = 1.375\) --- ### **Iteration 4** \[ x_{mid} = \frac{1.25 + 1.375}{2} = 1.3125 \] Calculate \(f(1.3125)\): \[ f(1.3125) = (1.3125)^3 - 1.3125 - 1 \\ = 2.258 - 1.3125 - 1 = -0.0545 \] Since \(f(1.3125) < 0\), root between 1.3125 and 1.375. Update: - \(x_{low} = 1.3125\) - \(x_{high} = 1.375\) --- ### **Iteration 5** \[ x_{mid} = \frac{1.3125 + 1.375}{2} = 1.34375 \] Calculate \(f(1.34375)\): \[ f(1.34375) = (1.34375)^3 - 1.34375 - 1 \\ = 2.429 - 1.34375 - 1 = 0.085 \] Since \(f(1.34375) > 0\), root between 1.3125 and 1.34375. Update: - \(x_{low} = 1.3125\) - \(x_{high} = 1.34375\) --- ### **Iteration 6** \[ x_{mid} = \frac{1.3125 + 1.34375}{2} = 1.328125 \] Calculate \(f(1.328125)\): \[ f(1.328125) = (1.328125)^3 - 1.328125 - 1 \\ = 2.342 - 1.328125 - 1 = 0.0139 \] Since \(f(1.328125) > 0\), root between 1.3125 and 1.328125. Update: - \(x_{low} = 1.3125\) - \(x_{high} = 1.328125\) --- ### **Final approximate root after 6 iterations:** \[ x_{mid} = 1.3281 \quad (\text{to 4 decimal places}) \] --- ### **Final Answer:** \[ \boxed{ \text{Root} \approx \mathbf{1.3281} } \] --- Let me know if you'd like further clarification or additional iterations!

Certainly! Here is a detailed solution for **Question 4**, parts (a), (b), and (c): --- ## Question 4 ### (a) Find the Taylor series of \(f(x) = \ln x\) centered at \(x_0 = 1\). ### **Solution:** The Taylor series expansion of a function \(f(x)\) about \(x_0=1\) is: \[ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!} (x-1)^n \] **Step 1: Find derivatives of \(f(x) = \ln x\):** \[ f(x) = \ln x \] \[ f'(x) = \frac{1}{x} \] \[ f''(x) = -\frac{1}{x^2} \] \[ f'''(x) = 2 \times \frac{1}{x^3} = \frac{2}{x^3} \] \[ f^{(4)}(x) = -6 \times \frac{1}{x^4} = -\frac{6}{x^4} \] In general, the \(n\)-th derivative: \[ f^{(n)}(x) = (-1)^{n-1} (n-1)! \times \frac{1}{x^n} \] **Step 2: Evaluate derivatives at \(x=1\):** \[ f^{(n)}(1) = (-1)^{n-1} (n-1)! \times 1 = (-1)^{n-1} (n-1)! \] **Step 3: Write the series:** \[ f(x) = \sum_{n=1}^\infty \frac{f^{(n)}(1)}{n!} (x-1)^n \] \[ = \sum_{n=1}^\infty \frac{(-1)^{n-1} (n-1)!}{n!} (x-1)^n \] Note: \[ \frac{(n-1)!}{n!} = \frac{1}{n} \] So, \[ f(x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n \] **Answer:** \[ \boxed{ \ln x = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n \quad \text{for } |x-1| < 1 } \] --- ### (b) Find the Maclaurin series for \(f(x) = \sin x\). ### **Solution:** The Maclaurin series (Taylor series at \(x=0\)) is: \[ \sin x = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k \] **Step 1: Derivatives of \(\sin x\):** \[ f(x) = \sin x \] \[ f'(x) = \cos x \] \[ f''(x) = -\sin x \] \[ f'''(x) = -\cos x \] \[ f^{(4)}(x) = \sin x \] The derivatives cycle every 4 steps. **Step 2: Evaluate at \(x=0\):** \[ f(0) = \sin 0 = 0 \] \[ f'(0) = \cos 0 = 1 \] \[ f''(0) = -\sin 0 = 0 \] \[ f'''(0) = -\cos 0 = -1 \] \[ f^{(4)}(0) = \sin 0 = 0 \] and so on. **Step 3: Write the series:** Only derivatives at odd orders are non-zero: - For odd \(k=2m+1\): \[ f^{(2m+1)}(0) = (-1)^m \] since: \[ f^{(1)}(0) = 1, \quad f^{(3)}(0) = -1, \quad f^{(5)}(0) = 1, \quad f^{(7)}(0) = -1, \dots \] **Step 4: Final series:** \[ \sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} \] **Answer:** \[ \boxed{ \sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} } \] --- ### (c) Determine for which \(x\)-values the series \(\sum_{n=0}^\infty \frac{(2x)^n}{n!}\) converges, specifically at \(x=1\). ### **Solution:** The series: \[ \sum_{n=0}^\infty \frac{(2x)^n}{n!} \] is the power series expansion of the exponential function: \[ e^{2x} = \sum_{n=0}^\infty \frac{(2x)^n}{n!} \] **Step 1: Convergence for all real \(x\):** The exponential function \(e^{2x}\) is entire, meaning its power series converges **for all real \(x\)**. **Step 2: At \(x=1\):** \[ \sum_{n=0}^\infty \frac{(2 \times 1)^n}{n!} = \sum_{n=0}^\infty \frac{2^n}{n!} = e^{2} \] which converges. --- ## **Summary:** | Part | Result | Convergence domain | |--------|---------|---------------------| | (a) | \(\ln x = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n\) | \(|x-1| < 1\) | | (b) | \(\sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\) | All \(x\) | | (c) | Series converges for all real \(x\), including \(x=1\) | All real \(x\) | --- Let me know if you'd like me to clarify any part further!

Certainly! Here are all the final answers summarized: --- **Question 2:** \(\boxed{2.512}\) --- **Question 3:** \(\boxed{1.3281}\) --- **Question 4:** - (a) \(\ln x = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n\) (valid for \(|x-1| < 1\)) - (b) \(\sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\) (valid for all \(x\)) - (c) The series \(\sum_{n=0}^\infty \frac{(2x)^n}{n!}\) converges for all real \(x\), including at \(x=1\). ---