Let's break down the problem step-by-step as requested. ### Given: \[ \int_0^1 23 \cos(x^2) \, dx \] We'll use the **Trapezoidal Rule** (\(T_8\)) and **Midpoint Rule** (\(M_8\)) with \(n = 8\) subintervals. --- ## (a) Find the approximations \(T_8\) and \(M_8\) ### **Step 1: Set up subintervals and required values** - \(a = 0\), \(b = 1\), \(n = 8\) - \(h = \frac{b-a}{n} = \frac{1-0}{8} = 0.125\) - \(f(x) = 23\cos(x^2)\) ### **Step 2: Calculate the Trapezoidal and Midpoint Approximations** #### **Trapezoidal Rule:** \[ T_8 = \frac{h}{2} \left[f(x_0) + 2\sum_{i=1}^{n-1} f(x_i) + f(x_n)\right] \] where \(x_i = a + i h\). #### **Midpoint Rule:** \[ M_8 = h \sum_{i=1}^{n} f\left(a + \left(i-\frac{1}{2}\right)h\right) \] Now let's compute the values: #### **Trapezoidal Rule Points:** \(x_0 = 0\), \(x_1 = 0.125\), \(x_2 = 0.25\), ..., \(x_8 = 1\) Calculate \(f(x_i)\) for each \(i\): - \(f(0) = 23\cos(0) = 23\) - \(f(0.125) = 23\cos(0.125^2) = 23\cos(0.015625)\) - \(f(0.25) = 23\cos(0.25^2) = 23\cos(0.0625)\) - ... - \(f(1) = 23\cos(1)\) Compute each value (rounded to 6 decimals): \[ \begin{align*} f(0) &= 23.000000 \\ f(0.125) &= 23 \times \cos(0.015625) = 23 \times 0.999877 = 22.997171 \\ f(0.25) &= 23 \times \cos(0.0625) = 23 \times 0.998046 = 22.955058 \\ f(0.375) &= 23 \times \cos(0.140625) = 23 \times 0.990125 = 22.773875 \\ f(0.5) &= 23 \times \cos(0.25) = 23 \times 0.968912 = 22.284976 \\ f(0.625) &= 23 \times \cos(0.390625) = 23 \times 0.924101 = 21.254323 \\ f(0.75) &= 23 \times \cos(0.5625) = 23 \times 0.846073 = 19.459679 \\ f(0.875) &= 23 \times \cos(0.765625) = 23 \times 0.723980 = 16.651538 \\ f(1) &= 23 \times \cos(1) = 23 \times 0.540302 = 12.426946 \\ \end{align*} \] Sum up as per the formula: \[ T_8 = \frac{0.125}{2} \left[23.000000 + 2(22.997171 + 22.955058 + 22.773875 + 22.284976 + 21.254323 + 19.459679 + 16.651538) + 12.426946\right] \] Compute the sum inside the brackets: \[ 2 \times (22.997171 + 22.955058 + 22.773875 + 22.284976 + 21.254323 + 19.459679 + 16.651538) = 2 \times 148.376620 = 296.753240 \] \[ 23.000000 + 296.753240 + 12.426946 = 332.180186 \] So, \[ T_8 = 0.0625 \times 332.180186 = 20.761261 \] #### **Midpoint Rule Points:** Midpoints: \(x_{0.5}, x_{1.5}, ..., x_{7.5}\) where \(x_{i-0.5} = (i-0.5)h\) - \(0.0625, 0.1875, 0.3125, 0.4375, 0.5625, 0.6875, 0.8125, 0.9375\) Calculate \(f\) at each midpoint: \[ \begin{align*} f(0.0625) &= 23 \times \cos(0.0625^2) = 23 \times \cos(0.00390625) = 22.999604 \\ f(0.1875) &= 23 \times \cos(0.1875^2) = 23 \times \cos(0.03515625) = 22.985679 \\ f(0.3125) &= 23 \times \cos(0.3125^2) = 23 \times \cos(0.09765625) = 22.889357 \\ f(0.4375) &= 23 \times \cos(0.4375^2) = 23 \times \cos(0.19140625) = 22.614472 \\ f(0.5625) &= 23 \times \cos(0.5625^2) = 23 \times \cos(0.31640625) = 22.070189 \\ f(0.6875) &= 23 \times \cos(0.6875^2) = 23 \times \cos(0.47265625) = 21.174324 \\ f(0.8125) &= 23 \times \cos(0.66015625) = 23 \times 0.789725 = 18.164810 \\ f(0.9375) &= 23 \times \cos(0.87890625) = 23 \times 0.637728 = 14.667748 \\ \end{align*} \] Sum: \[ 22.999604 + 22.985679 + 22.889357 + 22.614472 + 22.070189 + 21.174324 + 18.164810 + 14.667748 = 167.566183 \] So, \[ M_8 = 0.125 \times 167.566183 = 20.945773 \] --- ### **Summary for (a):** \[ \boxed{ \begin{align*} T_8 &= 20.761261 \\ M_8 &= 20.945773 \\ \end{align*} } \] --- ## (b) Estimate the errors in the approximations ### **Step 1: Error Bound Formulas** For the Trapezoidal Rule: \[ |E_T| \leq \frac{(b-a)^3}{12n^2} \max |f''(x)| \] For the Midpoint Rule: \[ |E_M| \leq \frac{(b-a)^3}{24n^2} \max |f''(x)| \] ### **Step 2: Find \(\max |f''(x)|\) and plug values** - \(b-a = 1\) - \(n = 8\) - \(f(x) = 23\cos(x^2)\) - \(f''(x) = -92x \sin(x^2) - 46 \cos(x^2)\) - The range of \(\sin\) and \(\cos\) is \([-1,1]\), so the maximum occurs at \(x=1\): - \(92x \leq 92\) - \(|-92x \sin(x^2)| \leq 92\) - \(|-46\cos(x^2)| \leq 46\) - So, the sum: \(92 + 46 = 138\) So, use \(\max |f''(x)| = 138\) \[ |E_T| \leq \frac{1^3}{12 \times 8^2} \times 138 = \frac{1}{12 \times 64} \times 138 = \frac{138}{768} = 0.1796875 \] \[ |E_M| \leq \frac{1^3}{24 \times 8^2} \times 138 = \frac{138}{1536} = 0.0898438 \] --- ### **Summary for (b):** \[ \boxed{ \begin{align*} |E_T| &\leq 0.1796875 \\ |E_M| &\leq 0.0898438 \\ \end{align*} } \] (Round to 7 decimal places if needed.) --- ## (c) Find \(n\) for error less than 0.0001 ### **Step 1: Set up inequalities** For Trapezoidal: \[ \frac{1}{12n^2} \times 138 \leq 0.0001 \implies n^2 \geq \frac{138}{0.0001 \times 12} \] For Midpoint: \[ \frac{1}{24n^2} \times 138 \leq 0.0001 \implies n^2 \geq \frac{138}{0.0001 \times 24} \] ### **Step 2: Solve** - Trapezoidal: \[ n^2 \geq \frac{138}{0.0012} = 115000 \implies n \geq 339.1165 \] - Midpoint: \[ n^2 \geq \frac{138}{0.0024} = 57500 \implies n \geq 239.0457 \] --- ### **Summary for (c):** \[ \boxed{ \begin{align*} n &\geq 340 \quad \text{for } T_n \\ n &\geq 240 \quad \text{for } M_n \\ \end{align*} } \] --- **Let me know if you want the calculations for each step written out in more detail or need the answers in a specific format!**

Let's carefully solve the problem step-by-step. --- ### **Given integral:** \[ \int_0^1 23 \cos(x^2) \, dx \] We'll approximate this integral using the **Trapezoidal Rule (T₈)** and **Midpoint Rule (M₈)** with \(n=8\) subintervals. --- ## (a) Find the approximations \(T_8\) and \(M_8\) ### **Step 1: Determine \(h\) and points** - \(a=0\), \(b=1\) - \(n=8\) - Subinterval width: \[ h = \frac{b - a}{n} = \frac{1}{8} = 0.125 \] ### **Step 2: Compute \(f(x) = 23 \cos(x^2)\) at required points** - \(x_i = a + i h\) **At endpoints:** \[ f(0) = 23 \times \cos(0) = 23 \times 1 = 23 \] \[ f(1) = 23 \times \cos(1) \approx 23 \times 0.540302 = 12.426946 \] **At intermediate points:** | \(x_i\) | \(x_i^2\) | \(\cos(x_i^2)\) | \(f(x_i) = 23 \times \cos(x_i^2)\) | |---------|-----------|----------------|-----------------------------------| | 0.125 | 0.015625 | \(\cos(0.015625) \approx 0.999877\) | \(23 \times 0.999877 \approx 22.997171\) | | 0.25 | 0.0625 | \(\cos(0.0625) \approx 0.998046\) | \(23 \times 0.998046 \approx 22.955058\) | | 0.375 | 0.140625 | \(\cos(0.140625) \approx 0.990125\) | \(23 \times 0.990125 \approx 22.773875\) | | 0.5 | 0.25 | \(\cos(0.25) \approx 0.968912\) | \(23 \times 0.968912 \approx 22.284976\) | | 0.625 | 0.390625 | \(\cos(0.390625) \approx 0.924101\) | \(23 \times 0.924101 \approx 21.254323\) | | 0.75 | 0.5625 | \(\cos(0.5625) \approx 0.846073\) | \(23 \times 0.846073 \approx 19.459679\) | | 0.875 | 0.765625 | \(\cos(0.765625) \approx 0.722\) | \(23 \times 0.722 \approx 16.612\) | --- ### **Step 3: Compute \(T_8\) (Trapezoidal Rule)** \[ T_8 = \frac{h}{2} \left[ f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right] \] Sum of interior points: \[ f(0.125) + f(0.25) + f(0.375) + f(0.5) + f(0.625) + f(0.75) + f(0.875) \approx 22.997171 + 22.955058 + 22.773875 + 22.284976 + 21.254323 + 19.459679 + 16.612 \approx 148.337 \] Calculate: \[ T_8 = \frac{0.125}{2} [23 + 2 \times 148.337 + 12.426946] \] \[ = 0.0625 [23 + 296.674 + 12.426946] = 0.0625 \times 332.100946 \approx 20.756935 \] --- ### **Step 4: Compute \(M_8\) (Midpoint Rule)** Midpoints: \[ x_{i-\frac{1}{2}} = \left(i - \frac{1}{2}\right) h \] for \(i=1 \text{ to } 8\): | \(i\) | \(x_{i-\frac{1}{2}}\) | \(x^2\) | \(f(x_{i-\frac{1}{2}})\) | |-------|------------------------|---------|--------------------------| | 1 | 0.0625 | 0.00390625 | \(\approx 23 \times 0.999977 \approx 22.9996\) | | 2 | 0.1875 | 0.03515625 | \(\approx 23 \times 0.998046 \approx 22.9551\) | | 3 | 0.3125 | 0.09765625 | \(\approx 23 \times 0.990125 \approx 22.7739\) | | 4 | 0.4375 | 0.19140625 | \(\approx 23 \times 0.9814 \approx 22.5742\) | | 5 | 0.5625 | 0.31640625 | \(\approx 23 \times 0.9486 \approx 21.8178\) | | 6 | 0.6875 | 0.47265625 | \(\approx 23 \times 0.8789 \approx 20.209\) | | 7 | 0.8125 | 0.66015625 | \(\approx 23 \times 0.7897 \approx 18.164\) | | 8 | 0.9375 | 0.87890625 | \(\approx 23 \times 0.6377 \approx 14.667\) | Sum: \[ \approx 22.9996 + 22.9551 + 22.7739 + 22.5742 + 21.8178 + 20.209 + 18.164 + 14.667 \approx 166.1616 \] Calculate: \[ M_8 = h \times \text{sum} = 0.125 \times 166.1616 \approx 20.7702 \] --- ## **Final approximations:** \[ \boxed{ \begin{aligned} T_8 &\approx \mathbf{20.756935} \\ M_8 &\approx \mathbf{20.7702} \end{aligned} } \] --- ## (b) Error bounds estimate ### **Step 1: Error bounds formulas** - **Trapezoidal:** \(\left|E_T\right| \leq \frac{(b - a)^3}{12 n^2} \max |f''(x)|\) - **Midpoint:** \(\left|E_M\right| \leq \frac{(b - a)^3}{24 n^2} \max |f''(x)|\) Given \(b-a=1\), \(n=8\), and the maximum of \(|f''(x)|\). ### **Step 2: Estimate \(\max |f''(x)|\)** - \(f(x) = 23 \cos(x^2)\) - \(f''(x) = -92x \sin(x^2) - 46 \cos(x^2)\) Range considerations: - \(|\sin(x^2)| \leq 1\) - \(|\cos(x^2)| \leq 1\) Maximum of \(|f''(x)|\): \[ |f''(x)| \leq 92|x| \times 1 + 46 \times 1 = 92|x| + 46 \] On \([0,1]\), maximum at \(x=1\): \[ |f''(x)| \leq 92 \times 1 + 46 = 138 \] ### **Step 3: Compute error bounds** \[ |E_T| \leq \frac{1^3}{12 \times 8^2} \times 138 = \frac{138}{768} \approx 0.1796875 \] \[ |E_M| \leq \frac{1^3}{24 \times 8^2} \times 138 = \frac{138}{1536} \approx 0.0898438 \] --- ## **(c) Determining \(n\) for desired accuracy (\(<0.00017\))** Set the error bounds: ### **For the Trapezoidal Rule:** \[ \frac{(b - a)^3}{12 n^2} \times 138 \leq 0.00017 \] \[ \frac{138}{12 n^2} \leq 0.00017 \] \[ n^2 \geq \frac{138}{12 \times 0.00017} \approx \frac{138}{0.00204} \approx 67647.1 \] \[ n \geq \sqrt{67647.1} \approx 260 \] ### **For the Midpoint Rule:** \[ \frac{(b - a)^3}{24 n^2} \times 138 \leq 0.00017 \] \[ \frac{138}{24 n^2} \leq 0.00017 \] \[ n^2 \geq \frac{138}{24 \times 0.00017} \approx \frac{138}{0.00408} \approx 33824.5 \] \[ n \geq \sqrt{33824.5} \approx 184 \] --- ## **Final answers:** ### **(a)** \[ \boxed{ T_8 \approx 20.756935, \quad M_8 \approx 20.7702 } \] ### **(b)** \[ \boxed{ |E_T| \leq 0.1796875, \quad |E_M| \leq 0.0898438 } \] ### **(c)** To achieve accuracy within \(0.00017\): \[ n \geq \boxed{260} \quad \text{for } T_n \] \[ n \geq \boxed{184} \quad \text{for } M_n \] --- If further clarification is needed, feel free to ask!