Q. Dividing the interval $[0,6]$ into $6$ equal parts and by using trapezoidal rule the value of $\int\limits_{0}^{6} x^{3} d x$ is approximately :
EAMCETEAMCET 2006
Solution:
$x$
0
1
2
3
4
5
6
$f (x)$
0
1
8
27
64
125
216
By trapezoidal rule
$\int\limits_{0}^{6} x^{3} d x=\frac{h}{2}\left\{y_{0}+y_{6}+2\left(y_{1}+y_{2}\right.\right.\left. \left.+y_{3}+y_{4}+y_{5}\right)\right\} $
$= \frac{1}{2}\{0+216+2(1+8+27+64+125)\} $
$= \frac{1}{2}\{216+450\}=\frac{666}{2}$
$= 333$
| $x$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| $f (x)$ | 0 | 1 | 8 | 27 | 64 | 125 | 216 |