Question

using the midpoint rule , approximate the integral:
integral sign b=pi, a=0 sec(x/3)dx , n=6
Thanx

Answer 1

if you carefully divide your interval, it should be
\[ x_i=\frac{i\pi}{6} \] for \[ i=0,1,2,3,4,5,6\]
Hence, the integral would become
\[ \int_0^{\pi} \sec\left(\frac{x}{3}\right)dx =\frac{\pi}{12}\left(\sum_{i=0}^5\sec\left(\frac{x_i+x_{i+1}}{2}\right)\right)\]

Answer 2

sorry, there is a typo, the thing in side sec on the right hand side should be divided by 6, not by 2.

Answer 3

but don't we need to find delta x first ?

Answer 4

delta x is \[ \frac{\pi}{12} \]

Answer 5

so let me see if I got this right .....pi-0/6= pi/ 12 ?? shouldn't it be pi/6?

Answer 6

Hang on, you are right, it should be pi/6, excuse me, I am on the rush

Answer 7

ok so the xi=xi+1 should be divided by 2 not 6 right ?

Answer 8

no, it is 6, indeed, you have totally 7 discrete points
\[ x_0=0, x_1=\frac{\pi}{6}, x_2=\frac{2\pi}{6}, x_3=\frac{3\pi}{6},x_4=\frac{4\pi}{6},x_5=\frac{5\pi}{6},x_6=\pi \]
and obviously, these 7 points will make up 6 intervals.

Answer 9

so I can just go ahead and plug in those points into the original function right ? or I can add them up and divide by 6?

Answer 10

yup, use the formula for the midpoint as I have given you, but it should be divided by 6, not by 2