Let's solve the problem step by step: --- ## **4(a) Write the indefinite integral as a Maclaurin series** Given the Maclaurin series for \(\sin(x)\): \[ \sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \] We want to find the Maclaurin series for: \[ \int \sin(x^4) \, dx \] **Step 1.** Substitute \(x^4\) for \(x\) in the series for \(\sin(x)\): \[ \sin(x^4) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^4)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4(2n+1)}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{8n+4}}{(2n+1)!} \] **Step 2.** Integrate term by term: \[ \int \sin(x^4) dx = \int \left[ \sum_{n=0}^{\infty} (-1)^n \frac{x^{8n+4}}{(2n+1)!} \right] dx = \sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)!} \int x^{8n+4} dx \] \[ = \sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)!} \cdot \frac{x^{8n+5}}{8n+5} + C \] \[ = C + \sum_{n=0}^{\infty} (-1)^n \frac{x^{8n+5}}{(2n+1)!(8n+5)} \] --- **List the first 5 terms:** For \(n=0\): \[ (-1)^0 \frac{x^{5}}{1!\cdot 5} = \frac{x^5}{5} \] For \(n=1\): \[ (-1)^1 \frac{x^{13}}{3!\cdot 13} = -\frac{x^{13}}{6 \cdot 13} = -\frac{x^{13}}{78} \] For \(n=2\): \[ (-1)^2 \frac{x^{21}}{5! \cdot 21} = \frac{x^{21}}{120 \cdot 21} = \frac{x^{21}}{2520} \] For \(n=3\): \[ (-1)^3 \frac{x^{29}}{7! \cdot 29} = -\frac{x^{29}}{5040 \cdot 29} = -\frac{x^{29}}{146160} \] For \(n=4\): \[ (-1)^4 \frac{x^{37}}{9! \cdot 37} = \frac{x^{37}}{362880 \cdot 37} = \frac{x^{37}}{13426560} \] So, the first five terms are: \[ \boxed{ \int \sin(x^4) dx = C + \frac{x^5}{5} - \frac{x^{13}}{78} + \frac{x^{21}}{2520} - \frac{x^{29}}{146160} + \frac{x^{37}}{13426560} + \cdots } \] --- ## **4(b) Use the first 5 terms to approximate \(\int_0^1 \sin(x^4) dx\) (round to 6 decimal places)** \[ \int_0^1 \sin(x^4) dx \approx \left[ \frac{x^5}{5} - \frac{x^{13}}{78} + \frac{x^{21}}{2520} - \frac{x^{29}}{146160} + \frac{x^{37}}{13426560} \right]_0^1 \] Plug in \(x = 1\): \[ = \frac{1}{5} - \frac{1}{78} + \frac{1}{2520} - \frac{1}{146160} + \frac{1}{13426560} \] Calculate each term: - \(\frac{1}{5} = 0.200000\) - \(\frac{1}{78} \approx 0.012821\) - \(\frac{1}{2520} \approx 0.000397\) - \(\frac{1}{146160} \approx 0.00000684\) - \(\frac{1}{13426560} \approx 0.0000000745\) Sum: \[ 0.200000 - 0.012821 + 0.000397 - 0.00000684 + 0.0000000745 \] \[ = 0.200000 - 0.012821 = 0.187179 \] \[ 0.187179 + 0.000397 = 0.187576 \] \[ 0.187576 - 0.00000684 = 0.18756916 \] \[ 0.18756916 + 0.0000000745 \approx 0.18756923 \] **Final answer (rounded to 6 decimal places):** \[ \boxed{0.187569} \] --- ## **Summary of Answers:** **(a)** \(\boxed{C + \frac{x^5}{5} - \frac{x^{13}}{78} + \frac{x^{21}}{2520} - \frac{x^{29}}{146160} + \frac{x^{37}}{13426560} + \cdots}\) **(b)** \(\boxed{0.187569}\)

Certainly! Here's a detailed, step-by-step explanation with theoretical insights, detailed calculations, and the final answers: --- ## **Part 4(a): Expressing the indefinite integral as a Maclaurin series** ### **Step 1: Recall the Maclaurin series for \(\sin x\):** \[ \sin x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \] **This series converges for all real \(x\).** ### **Step 2: Substitute \(x^4\) into the series for \(\sin x\):** We want the series for \(\sin(x^4)\): \[ \sin(x^4) = \sum_{n=0}^\infty (-1)^n \frac{(x^4)^{2n+1}}{(2n+1)!} \] Calculate \((x^4)^{2n+1} = x^{4(2n+1)} = x^{8n+4}\). Thus, \[ \sin(x^4) = \sum_{n=0}^\infty (-1)^n \frac{x^{8n+4}}{(2n+1)!} \] ### **Step 3: Integrate term-by-term:** The indefinite integral: \[ \int \sin(x^4) dx = \int \left[\sum_{n=0}^\infty (-1)^n \frac{x^{8n+4}}{(2n+1)!}\right] dx \] Interchanging sum and integral (valid within the radius of convergence): \[ = \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} \int x^{8n+4} dx \] The integral: \[ \int x^{8n+4} dx = \frac{x^{8n+5}}{8n+5} + C \] Therefore, the Maclaurin series for the indefinite integral is: \[ \boxed{ \int \sin(x^4) dx = C + \sum_{n=0}^\infty (-1)^n \frac{x^{8n+5}}{(2n+1)! (8n+5)} } \] --- ### **First 5 terms:** Calculate for \(n=0,1,2,3,4\): | \(n\) | Term expression | Numerical form | |-------|-------------------|----------------| | 0 | \(\frac{x^5}{1! \times 5} = \frac{x^5}{5}\) | \(\frac{x^5}{5}\) | | 1 | \(-\frac{x^{13}}{3! \times 13} = -\frac{x^{13}}{6 \times 13} = -\frac{x^{13}}{78}\) | \(-\frac{x^{13}}{78}\) | | 2 | \(\frac{x^{21}}{5! \times 21} = \frac{x^{21}}{120 \times 21} = \frac{x^{21}}{2520}\) | \(\frac{x^{21}}{2520}\) | | 3 | \(-\frac{x^{29}}{7! \times 29} = -\frac{x^{29}}{5040 \times 29} = -\frac{x^{29}}{146160}\) | \(-\frac{x^{29}}{146160}\) | | 4 | \(\frac{x^{37}}{9! \times 37} = \frac{x^{37}}{362880 \times 37} = \frac{x^{37}}{13426560}\) | \(\frac{x^{37}}{13426560}\) | **Final Series:** \[ \boxed{ \int \sin(x^4) dx \approx C + \frac{x^5}{5} - \frac{x^{13}}{78} + \frac{x^{21}}{2520} - \frac{x^{29}}{146160} + \frac{x^{37}}{13426560} } \] --- ## **Part 4(b): Approximate \(\int_0^1 \sin(x^4) dx\) using the first 5 terms** ### **Step 1: Evaluate the series at \(x=1\):** \[ \int_0^1 \sin(x^4) dx \approx \left[\frac{x^5}{5} - \frac{x^{13}}{78} + \frac{x^{21}}{2520} - \frac{x^{29}}{146160} + \frac{x^{37}}{13426560}\right]_0^1 \] Plugging in \(x=1\): \[ = \frac{1}{5} - \frac{1}{78} + \frac{1}{2520} - \frac{1}{146160} + \frac{1}{13426560} \] ### **Step 2: Calculate each term:** - \(\frac{1}{5} = 0.200000\) - \(\frac{1}{78} \approx 0.0128205128\) - \(\frac{1}{2520} \approx 0.0003968254\) - \(\frac{1}{146160} \approx 0.0000068446\) - \(\frac{1}{13426560} \approx 0.0000000744\) ### **Step 3: Sum the terms:** \[ 0.200000 - 0.0128205128 + 0.0003968254 - 0.0000068446 + 0.0000000744 \] Step-by-step: - \(0.200000 - 0.0128205128 = 0.1871794872\) - \(0.1871794872 + 0.0003968254 = 0.1875763126\) - \(0.1875763126 - 0.0000068446 = 0.187569468\) - \(0.187569468 + 0.0000000744 \approx 0.1875695424\) **Rounded to six decimal places:** \[ \boxed{ \boxed{0.187569} } \] --- ## **Summary:** - **(a)** The indefinite integral as a Maclaurin series: \[ \boxed{ \int \sin(x^4) dx = C + \frac{x^5}{5} - \frac{x^{13}}{78} + \frac{x^{21}}{2520} - \frac{x^{29}}{146160} + \frac{x^{37}}{13426560} + \cdots } \] - **(b)** Approximate value of \(\int_0^1 \sin(x^4) dx\): \[ \boxed{ 0.187569 } \] --- **Note:** The series converges rapidly for small \(x\), and using the first five terms yields a good approximation within six decimal places.

1. The Maclaurin series expansion expresses a function as an infinite sum of its derivatives at zero, enabling approximation of complicated functions through polynomial terms. 2. Integrating term-by-term in the series allows us to find the series for the integral of the function, facilitating approximation of definite integrals using partial sums.