For the integral ${\int }_0^{\pi /2}\left(8+4\cos x\right)dx,$ the absolute percentage error in numerical evaluation with the Trapezoidal rule, using only the endpoints, is ______ (round off to one decimal place).

Correct answer is : 5.2

Given, Number of interval = 1

a = 0,  $\mathrm{b}=\frac{\pi }{2}$

$\begin{array}{l}\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{l}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{h}}\\ \Rightarrow 1=\frac{\frac{\pi }{2}-0}{\mathrm{h}}\\ \therefore \mathrm{h}=\frac{\pi }{2}\end{array}$

True value

${\int }_0^{\pi /2}\left(8+4\cos x\right)dx={\left[8x+4sinx\right]}_0^{\frac{\pi }{2}}$

$\Rightarrow \mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}=\left[\left(8\times \frac{\pi }{2}\right)+\left(4\times \sin \left(\frac{\pi }{2}\right)\right)\right]$

⇒ True value = 4π + 4 = 16.56

Approximate value

By trapezoidal rule

$h=\frac{\left(\frac{\pi }{2}-0\right)}{1}=\frac{\pi }{2}$

According to the trapezoidal rule

$\begin{array}{l}\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\frac{\mathrm{h}}{2}\left[{\mathrm{y}}_{\mathrm{o}}+{\mathrm{y}}_{\mathrm{n}}+2\left({\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3\dots \right)\right]\\ \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(8+4\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}\right)\mathrm{d}\mathrm{x}=\frac{\pi }{4}\left(12+8\right)\end{array}$

⇒  Approximate value = 5π = 15.70

Now,

$\begin{array}{l}Absolutepercentageerror=\frac{\left|truevalue-approximatevalue\right|}{truevalue}\\ Absolutepercentageerror=\frac{\left|16.56-15.70\right|}{16.56}\times 100\end{array}$

∴ Absolute percentage error = 5.19 % ≈ 5.2 %