7 - use series to approximate the value of ∫₀...

# Answer: We first recall the Maclaurin series for $\cos t$: $\cos t=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n)!}t^{2n}=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+\cdots$. Let $t = x^{2}$, then $\cos(x^{2})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}(x^{2})^{2n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{4n}=1-\frac{x^{4}}{2!}+\frac{x^{8}}{4!}-\frac{x^{12}}{6!}+\cdots$. Now we integrate term - by - term: \[ \begin{align*} \int_{0}^{1}\cos(x^{2})dx&=\int_{0}^{1}\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{4n}dx\\ &=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}\int_{0}^{1}x^{4n}dx\\ &=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!(4n + 1)} \end{align*} \] This is an alternating series of the form $\sum_{n = 0}^{\infty}(-1)^{n}a_{n}$ where $a_{n}=\frac{1}{(2n)!(4n + 1)}$ and $a_{n+1}<a_{n}$ for all $n\geq0$ and $\lim_{n\rightarrow\infty}a_{n}=0$. The error bound for an alternating series $\sum_{n = 0}^{\infty}(-1)^{n}a_{n}$ is given by $|E_N|\leq a_{N + 1}$. We want $a_{N+1}<\frac{1}{100}$. For $n = 0$: $a_{0}=\frac{1}{(0)!(0 + 1)}=1$ For $n = 1$: $a_{1}=\frac{1}{2!(4\times1+1)}=\frac{1}{2\times5}=\frac{1}{10}$ For $n = 2$: $a_{2}=\frac{1}{4!(4\times2 + 1)}=\frac{1}{24\times9}=\frac{1}{216}<\frac{1}{100}$ So, $\int_{0}^{1}\cos(x^{2})dx\approx\sum_{n = 0}^{1}\frac{(-1)^{n}}{(2n)!(4n + 1)}=1-\frac{1}{10}=\frac{9}{10} = 0.9$ # Explanation: ## Step1: Recall Maclaurin series of $\cos t$ $\cos t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}t^{2n}$ ## Step2: Substitute $t=x^{2}$ $\cos(x^{2})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{4n}$ ## Step3: Integrate term - by - term $\int_{0}^{1}\cos(x^{2})dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!(4n + 1)}$ ## Step4: Apply error bound for alternating series Find $N$ such that $a_{N + 1}<\frac{1}{100}$ ## Step5: Calculate approximation $\int_{0}^{1}\cos(x^{2})dx\approx1-\frac{1}{10}=0.9$