The value of the integral $\int_0^4 \frac{d x}{1+x^2}$ obtained by using Trapezoidal rule with $h=1$ is

Question

Aicharya · Accepted Answer

Given integration is $\int_0^4 \frac{d x}{1+x^2}$ and $h=1$.

By using Trapezoidal rule, $\begin{aligned} \int_0^4 f(x) d x & =\frac{h}{2}\left[\left(y_0+y_4\right)+2\left(y_1+y_2+y_3\right)\right] \\ & =\frac{1}{2}\left[\left(1+\frac{1}{17}\right)+2\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)\right] \\ & =\frac{1}{2}\left[\frac{18}{17}+2\left(\frac{5+2+1}{10}\right)\right]\end{aligned}$ $\begin{aligned} & =\frac{1}{2}\left(\frac{18}{17}+\frac{8}{5}\right)=\frac{1}{2}\left(\frac{90+136}{85}\right) \\ & =\frac{1}{2}\left(\frac{226}{85}\right)=\frac{113}{85}\end{aligned}$

The value of the integral $\int_0^4 \frac{d x}{1+x^2}$ obtained by using Trapezoidal rule with $h=1$ is

Solution