MDoubt No 000020 - GATE

GATE [ME, 2016, 2 marks]
Question : The error in numerically computing the integral $\displaystyle\int\limits_0^{\pi}(\sin x+\cos x)dx$ using the Trapezoidal rule with three intervals of equal lenght between $0$ and $\pi$ is____.

Solution : Given $F(x)=\sin x+\cos x$
$h=\dfrac{b-a}{n}=\dfrac{\pi-0}{3}=\dfrac{\pi}{3}$

$x$	$0$	$\dfrac{\pi}{3}$	$\dfrac{2\pi}{3}$	$\pi$
$\sin x+\cos x$	$1$	$1.366$	$0.366$	$-1$

By Trapezoidal rule
$\displaystyle\int\limits_{a}^{b}F(x)dx=\dfrac{h}{2}[(y_0+y_{n})+2(y_1+y_2+\cdots+y_{n-1})]$
$\displaystyle\int\limits_{0}^{\pi}\sin x+\cos xdx=\dfrac{\pi/3}{2}[(1-1)+2(1.366+0.366)]$
$\displaystyle\int\limits_{0}^{\pi}\sin x+\cos xdx=1.181$
By definite integration
$\displaystyle\int\limits_{0}^{\pi}\sin x+\cos xdx=\left|-\cos x+\sin x\right|^{0}_{\pi}$
$\displaystyle\int\limits_{0}^{\pi}\sin x+\cos xdx=2$
Error $=$ Exact Value $-$ Approximate Value
$=2-1.181$
$=0.187$
CORRECT ANSWER : 0.187