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.

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

Short answer

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

Step-by-step solution

Step 1. Given information is: 

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

Step 2. Approximating the values 

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