Let's solve **part (a)** step by step using the **Composite Trapezoidal Rule**. ## Problem Approximate: \[ \int_1^2 x \ln x \, dx,\quad n = 4 \] --- ## Step 1: **Set up the Trapezoidal Rule** The composite trapezoidal rule for \( n \) subintervals over \([a, b]\): \[ \int_a^b f(x) dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n) \right] \] where \( h = \frac{b-a}{n} \) and \( x_i = a + ih \). --- ## Step 2: **Calculate \( h \) and the nodes \( x_i \)** Given: \( a = 1 \), \( b = 2 \), \( n = 4 \) \[ h = \frac{2-1}{4} = 0.25 \] So, - \( x_0 = 1 \) - \( x_1 = 1 + 0.25 = 1.25 \) - \( x_2 = 1 + 2 \times 0.25 = 1.5 \) - \( x_3 = 1 + 3 \times 0.25 = 1.75 \) - \( x_4 = 2 \) --- ## Step 3: **Calculate \( f(x_i) = x_i \ln x_i \) at each node** \[ \begin{align*} f(x_0) &= 1 \cdot \ln(1) = 0 \\ f(x_1) &= 1.25 \cdot \ln(1.25) \approx 1.25 \cdot 0.2231 = 0.2789 \\ f(x_2) &= 1.5 \cdot \ln(1.5) \approx 1.5 \cdot 0.4055 = 0.6083 \\ f(x_3) &= 1.75 \cdot \ln(1.75) \approx 1.75 \cdot 0.5606 = 0.9810 \\ f(x_4) &= 2 \cdot \ln(2) \approx 2 \cdot 0.6931 = 1.3862 \\ \end{align*} \] --- ## Step 4: **Apply the formula** \[ \int_1^2 x \ln x\, dx \approx \frac{0.25}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plug in values: \[ = 0.125 \left[0 + 2(0.2789) + 2(0.6083) + 2(0.9810) + 1.3862 \right] \] \[ = 0.125 \left[0 + 0.5578 + 1.2166 + 1.9620 + 1.3862 \right] \] \[ = 0.125 \left[5.1226\right] \] \[ = 0.6403 \] --- ## **Final Answer** \[ \boxed{0.640} \] --- **Let me know if you want the solution to another part!**

Certainly! Let's proceed with **part (b)** using the **Composite Trapezoidal Rule**. --- ## Part (b): Approximate \(\int_2^4 x^2 e^x \, dx\) with \(n=4\) ### Step 1: Identify parameters - \(a = 2\) - \(b = 4\) - \(n = 4\) Calculate \(h\): \[ h = \frac{b - a}{n} = \frac{4 - 2}{4} = 0.5 \] ### Step 2: Compute nodes \(x_i\): \[ x_0 = 2 \\ x_1 = 2 + 0.5 = 2.5 \\ x_2 = 3.0 \\ x_3 = 3.5 \\ x_4 = 4.0 \] ### Step 3: Evaluate \(f(x) = x^2 e^x\) at each node: - \(f(2) = 2^2 \times e^2 = 4 \times 7.3891 \approx 29.5564\) - \(f(2.5) = (2.5)^2 \times e^{2.5} = 6.25 \times 12.1825 \approx 76.1406\) - \(f(3) = 9 \times e^{3} = 9 \times 20.0855 \approx 180.7695\) - \(f(3.5) = 12.25 \times e^{3.5} = 12.25 \times 33.1155 \approx 406.9114\) - \(f(4) = 16 \times e^{4} = 16 \times 54.5982 \approx 873.5712\) ### Step 4: Apply the composite trapezoidal rule: \[ \int_2^4 x^2 e^x dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plugging in the values: \[ = \frac{0.5}{2} \left[ 29.5564 + 2(76.1406) + 2(180.7695) + 2(406.9114) + 873.5712 \right] \] \[ = 0.25 \left[ 29.5564 + 152.2812 + 361.539 + 813.8228 + 873.5712 \right] \] Sum inside brackets: \[ 29.5564 + 152.2812 = 181.8376 \\ 181.8376 + 361.539 = 543.3766 \\ 543.3766 + 813.8228 = 1357.1994 \\ 1357.1994 + 873.5712 = 2230.7706 \] Multiply by 0.25: \[ 0.25 \times 2230.7706 \approx 557.693 \] --- ## **Final Answer:** \[ \boxed{557.69} \] --- Would you like to continue with the next part?