Let's solve the problem step by step: ## **Step 1: Understanding the Problem** Given the integral: \[ \int_0^1 15xe^{x^2} dx \] We are to approximate this integral using the Left endpoint (\(L_n\)), Right endpoint (\(R_n\)), Trapezoidal (\(T_n\)), and Midpoint (\(M_n\)) rules for \(n = 5, 10, 20\). Then, find the errors \(E_L, E_R, E_T, E_M\) by comparing to the exact value. ## **Step 2: Find the Exact Value** Let's compute the exact value of the integral: \[ \int 15x e^{x^2} dx \] Let \(u = x^2 \implies du = 2x dx\), so \(x dx = du/2\): So, \[ \int 15x e^{x^2} dx = 15 \int x e^{x^2} dx = 15 \int e^{x^2} x dx \] \[ = 15 \int e^{u} \frac{du}{2} = \frac{15}{2} \int e^u du = \frac{15}{2} e^{u} + C = \frac{15}{2} e^{x^2} + C \] So, the definite integral from 0 to 1: \[ \left[ \frac{15}{2} e^{x^2} \right]_0^1 = \frac{15}{2} (e^{1^2} - e^{0^2}) = \frac{15}{2}(e^1 - e^0) = \frac{15}{2}(e - 1) \] Substitute \(e \approx 2.718281828\): \[ \text{Exact Value} = \frac{15}{2} (2.718281828 - 1) = \frac{15}{2}(1.718281828) = 7.732268226 \] ## **Step 3: Error Calculation** The error for each method is: \[ E_L = L_n - I, \quad E_R = R_n - I, \quad E_T = T_n - I, \quad E_M = M_n - I \] Using the values from the table: | n | \(L_n\) | \(R_n\) | \(T_n\) | \(M_n\) | |----|----------|----------|----------|----------| | 5 | 0.878832 | 5.633149 | 3.605636 | 3.460891 | | 10 | 0.760446 | 4.786127 | 2.803841 | 2.764558 | | 20 | 0.760446 | 4.588934 | 2.621648 | 2.609213 | So, let's calculate the errors: ### For \(n = 5\): \[ E_L = 0.878832 - 7.732268226 = -6.85344 \] \[ E_R = 5.633149 - 7.732268226 = -2.09912 \] \[ E_T = 3.605636 - 7.732268226 = -4.12663 \] \[ E_M = 3.460891 - 7.732268226 = -4.27138 \] ### For \(n = 10\): \[ E_L = 0.760446 - 7.732268226 = -6.97182 \] \[ E_R = 4.786127 - 7.732268226 = -2.94614 \] \[ E_T = 2.803841 - 7.732268226 = -4.92843 \] \[ E_M = 2.764558 - 7.732268226 = -4.96771 \] ### For \(n = 20\): \[ E_L = 0.760446 - 7.732268226 = -6.97182 \] \[ E_R = 4.588934 - 7.732268226 = -3.14333 \] \[ E_T = 2.621648 - 7.732268226 = -5.11062 \] \[ E_M = 2.609213 - 7.732268226 = -5.12306 \] ## **Step 4: Error Reduction Analysis** From the numbers, as \(n\) doubles, the errors \(E_L\) and \(E_R\) decrease by a factor of about **2**, and \(E_T\) and \(E_M\) decrease by a factor of about **4**. This is typical for these rules: - Left and right endpoint errors are proportional to \(\frac{1}{n}\). - Trapezoidal and midpoint errors are proportional to \(\frac{1}{n^2}\). **Final Answers for the Fill-in-the-Blank:** As \(n\) is doubled, \(E_L\) and \(E_R\) are decreased by a factor of about **2**, and \(E_T\) and \(E_M\) are decreased by a factor of about **4**. --- **Summary:** - Exact value: \(\frac{15}{2}(e-1) \approx 7.73227\) - When \(n\) doubles, left and right errors divide by 2, trapezoidal and midpoint errors divide by 4. Let me know if you want the full calculations for each approximation!