Correct Answer - Option 1 : 1.8277
By trapezoidal rule
\(\begin{array}{l} \mathop \smallint \limits_4^{5.2} {ln}x\ dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_6}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + {y_4}+{y_5}} \right)} \right]\\ \Rightarrow \mathop \smallint \limits_4^{5.2} {ln}xdx = \frac{{0.2}}{2}\left[ {\left( {1.39 + 1.65} \right) + 2\left( {1.44 + 1.48 + 1.53 + 1.57 + 1.61} \right)} \right]\\ \Rightarrow \mathop \smallint \limits_4^{5.2} {ln}xdx = 1.83 \end{array}\)
