Question

The value of the integral ${\int }_0^4\frac{dx}{1+x^2}$ obtained by using Trapezoidal rule with $h=1$ is

• A. $\frac{63}{85}$
• B. ${\tan }^{-1}\left(4\right)$
• C. $\frac{108}{85}$
• D. $\frac{113}{85}$

Answer 1

Answer: (D)

$\frac{113}{85}$

Let $f\left(x\right)=\frac{1}{1+x^2}$

We know that ${\int }_a^{b}f\left(x\right)=(b-a)\frac{\left[f\left(a\right)+f\left(b\right)\right]}{2}$

Given $h=1$

$\therefore {\int }_0^{4}f\left(x\right)dx={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx+{\int }_2^{3}f\left(x\right)dx+{\int }_3^{4}f\left(x\right)dx$

$=(1-0)\left[\frac{f\left(0\right)+f\left(1\right)}{2}\right]+(2-1)\left[\frac{f\left(1\right)+f\left(2\right)}{2}\right]+(3-2)\left[\frac{f\left(3\right)f\left(2\right)}{2}\right]+(4-3)\left[\frac{f\left(4\right)+f\left(3\right)}{2}\right]$

$=\frac{1}{2}[1+\frac{1}{2}+\frac{1}{2}+\frac{1}{5}+\frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{17}]$

$=\frac{170+27+255}{340}$

$=\frac{113}{85}$