Explanation:
(C) : We have, \(\int_{0}^{1} \frac{\mathrm{dx}}{1+\mathrm{x}}\)
Here, \(\mathrm{a}=0, \mathrm{~b}=1, \mathrm{n}=4\)
Now, \(\mathrm{nh}=\mathrm{b}-\mathrm{a}\)
\(4 \mathrm{~h}=1\)
\(\mathrm{~h}=\frac{1}{4}\)
The value of \(f(x)=\frac{1}{1+x}\) are given below
| $x$ | 0 | $\frac{1}{4}$ | $\frac{2}{4}$ | $\frac{3}{4}$ | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f(x)$ | $y_0=1$ | $y_1=\frac{4}{5}$ | $y_2=\frac{4}{6}$ | $y_3=\frac{4}{7}$ | $y_4=\frac{1}{2}$ |
By trapezoidal rule
\(\int_{0}^{1} \frac{\mathrm{dx}}{1+\mathrm{x}}=\frac{\mathrm{h}}{2}\left[\left(\mathrm{y}_{0}+\mathrm{y}_{4}\right)+2\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)\right]\)
\(\int_{0}^{1} \frac{\mathrm{dx}}{1+\mathrm{x}}=\frac{1}{8}\left[\left(1+\frac{1}{2}\right)+2\left(\frac{4}{5}+\frac{4}{6}+\frac{4}{7}\right)\right]\)
\(=\frac{1}{8}\left[\frac{3}{2}+8\left(\frac{42+35+30}{210}\right)\right]\)
\(=\frac{3}{16}+\frac{107}{210}\)
\(=0.1875+0.5095=0.6970\)