Let's solve each part step by step: --- ### 1. **Maclaurin Series for \(x^2 \cos x^2\)** **Maclaurin Series for \( \cos x \):** \[ \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \] **Substitute \( x^2 \) for \( x \):** \[ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(2n)!} \] **Multiply by \( x^2 \):** \[ x^2 \cos(x^2) = x^2 \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n)!} \] --- ### 2. **Express the Integral as a Power Series** We want to write: \[ \int_{0}^{1} x^2 \cos(x^2) \, dx \] Using the series: \[ \int_{0}^{1} x^2 \cos(x^2) \, dx = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n)!} dx \] We can integrate term by term: \[ = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \int_{0}^{1} x^{4n+2} dx \] \[ = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \left[ \frac{x^{4n+3}}{4n+3} \right]_{0}^{1} \] \[ = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(4n+3)} \] --- ### 3. **Approximate with the First Three Terms and Estimate the Error** #### **First Three Terms:** For \( n = 0, 1, 2 \): **n = 0:** \[ \frac{(-1)^0}{(0)!(3)} = \frac{1}{3} \] **n = 1:** \[ \frac{(-1)^1}{2!(7)} = -\frac{1}{2 \times 7} = -\frac{1}{14} \] **n = 2:** \[ \frac{(-1)^2}{4!(11)} = \frac{1}{24 \times 11} = \frac{1}{264} \] **Sum:** \[ \text{Approximation} = \frac{1}{3} - \frac{1}{14} + \frac{1}{264} \] #### **Error Estimate** The error is the magnitude of the next term (\(n = 3\)): \[ \text{Error} \leq \left| \frac{(-1)^3}{6!(15)} \right| = \frac{1}{720 \times 15} = \frac{1}{10800} \] --- ## **Final Answers:** ### 1. **Power series for \( x^2 \cos(x^2) \):** \[ x^2 \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n)!} \] ### 2. **Integral as a power series:** \[ \int_{0}^{1} x^2 \cos(x^2) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(4n+3)} \] ### 3. **First three term approximation and error:** \[ \text{Approximation} = \frac{1}{3} - \frac{1}{14} + \frac{1}{264} \] \[ \text{Error estimate:} \leq \frac{1}{10800} \]

Let's carefully perform each step with detailed calculations. --- ### **Part 1: Find the power series expansion of \( x^2 \cos x^2 \) using Maclaurin series** **Step 1:** Recall the Maclaurin series for \( \cos x \): \[ \cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \] **Step 2:** Substitute \( x^2 \) in place of \( x \): \[ \cos x^2 = \sum_{n=0}^\infty \frac{(-1)^n (x^2)^{2n}}{(2n)!} = \sum_{n=0}^\infty \frac{(-1)^n x^{4n}}{(2n)!} \] **Step 3:** Multiply by \( x^2 \): \[ x^2 \cos x^2 = x^2 \times \sum_{n=0}^\infty \frac{(-1)^n x^{4n}}{(2n)!} = \sum_{n=0}^\infty \frac{(-1)^n x^{4n + 2}}{(2n)!} \] **Result:** \[ \boxed{ x^2 \cos x^2 = \sum_{n=0}^\infty \frac{(-1)^n x^{4n+2}}{(2n)!} } \] --- ### **Part 2: Find the power series expansion of the integral** \[ I = \int_0^1 x^2 \cos x^2 \, dx \] **Step 1:** Substitute the series expansion: \[ I = \int_0^1 \left( \sum_{n=0}^\infty \frac{(-1)^n x^{4n+2}}{(2n)!} \right) dx \] **Step 2:** Interchange the sum and the integral (justified since the series converges uniformly on [0,1]): \[ I = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \int_0^1 x^{4n+2} dx \] **Step 3:** Compute the integral: \[ \int_0^1 x^{4n+2} dx = \left[ \frac{x^{4n+3}}{4n+3} \right]_0^1 = \frac{1}{4n+3} \] **Step 4:** Write the series for the integral: \[ I = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \times \frac{1}{4n+3} \] **Final Series:** \[ \boxed{ I = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)! \, (4n + 3)} } \] --- ### **Part 3: Approximate the integral using the first three terms and estimate the error** **Step 1:** Calculate the first three terms (\( n=0,1,2 \)): - **For \( n=0 \):** \[ \frac{(-1)^0}{(0)! \times 3} = \frac{1}{1 \times 3} = \frac{1}{3} \] - **For \( n=1 \):** \[ \frac{(-1)^1}{(2)! \times 7} = -\frac{1}{2 \times 7} = -\frac{1}{14} \] - **For \( n=2 \):** \[ \frac{(-1)^2}{(4)! \times 11} = \frac{1}{24 \times 11} = \frac{1}{264} \] **Step 2:** Sum these three terms: \[ \text{Approximate value} = \frac{1}{3} - \frac{1}{14} + \frac{1}{264} \] --- ### **Numerical calculation:** - Find common denominator for easier addition: \[ \text{Denominator} = 3 \times 14 \times 264 = \text{(but easier to convert each to decimal)} \] Alternatively, convert each to decimal: \[ \frac{1}{3} \approx 0.333333 \] \[ \frac{1}{14} \approx 0.071429 \] \[ \frac{1}{264} \approx 0.003788 \] Now, \[ I_{\text{approx}} \approx 0.333333 - 0.071429 + 0.003788 = 0.265692 \] --- ### **Error estimate:** The next term corresponds to \( n=3 \): \[ \frac{(-1)^3}{(6)! \times 15} = - \frac{1}{720 \times 15} = - \frac{1}{10800} \] The magnitude of the next term: \[ |T_{4}| = \frac{1}{10800} \approx 0.00009259 \] **Estimate of the error** (the remainder after 3 terms): \[ \boxed{ \text{Error} \leq 0.0000926 } \] --- ### **Final summarized answers:** - **Power series for \( x^2 \cos x^2 \):** \[ \boxed{ x^2 \cos x^2 = \sum_{n=0}^\infty \frac{(-1)^n x^{4n+2}}{(2n)!} } \] - **Integral as a series:** \[ \boxed{ \int_0^1 x^2 \cos x^2 dx = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)! (4n+3)} } \] - **Approximate value using first three terms:** \[ \boxed{ I \approx 0.265692 } \] - **Error estimate:** \[ \boxed{ \text{Error} \leq 0.0000926 } \]