Q. 66 - 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. 

${\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}$

Step-by-Step Solution

Answer

${\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}\approx 0.761$

Step 1. Given information is:

${\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}$

Step 2. Approximating the values

$$\begin{gathered} \mathrm{From} \mathrm{Q} 55. \\ {\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}=\frac{(\ln 4)}{2}+\sum _{\mathrm{k}=1}^{\infty } \frac{{(-1)}^{\mathrm{k}+1}}{\mathrm{k}.2^{2\mathrm{k}}}\frac{(1-0.5^{2\mathrm{k}+1})}{2\mathrm{k}+1} \\ \mathrm{This} \mathrm{implies} \mathrm{that} \mathrm{to} \mathrm{approximate} \mathrm{the} \mathrm{values} \mathrm{of} \mathrm{integrals}, \mathrm{put} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{k} \\ {\int }_{0.5}^1\ln (4+{\mathrm{x}}^2)\mathrm{dx}=0.693147+\frac{0.875}{4·3}-\frac{0.96875}{2·16·5}+... \\ =0.693147+0.0729-0.00605+... \\ \approx 0.761 \end{gathered}$$