Q. 69 - Calculus

Question

Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values. 

${\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}$

Step-by-Step Solution

Answer

${\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}\approx -0.481$

Step 1. Given information is:

${\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}$

Step 2. Approximating the values

$$\begin{gathered} \mathrm{From} \mathrm{Q} 59. \\ {\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{1-{\left(0.5\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right) \\ \mathrm{This} \mathrm{implies} \mathrm{that} \mathrm{to} \mathrm{approximate} \mathrm{the} \mathrm{values} \mathrm{of} \mathrm{integrals}, \mathrm{put} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{k} \\ {\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}=\frac{1}{3}\left(1-0.5^4\right)-\frac{3}{5}\left(1-0.5^6\right)+\frac{9}{2!·7}\left(1-0.5^8\right) \\ =0.3125-0.5906 + 0.6403-0.4995 \\ \approx -0.481 \end{gathered}$$