Values of the function $y=\frac{1}{1+x^{2}}$ are $y(0)=1 ; y(1)=0.5 ; y(2)=0.2 ; y(3)=0.1 ; y(4)=0.0588$
$\mathrm{y}(5)=0.0385 ; \mathrm{y}(6)=0.027 .$ Using Simpson's one-third rule, the value of the integral $\int_{0}^{6} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}$ is___________

Question ID - 156984 | SaraNextGen Top Answer

Values of the function $y=\frac{1}{1+x^{2}}$ are $y(0)=1 ; y(1)=0.5 ; y(2)=0.2 ; y(3)=0.1 ; y(4)=0.0588$
$\mathrm{y}(5)=0.0385 ; \mathrm{y}(6)=0.027 .$ Using Simpson's one-third rule, the value of the integral $\int_{0}^{6} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}$ is___________