Question

Calculate M6 for f(x)=4⋅ln(x^2) over [1,2].

Answer 1

In this we need to approximate definite integral by midpoint formula.
According to this formula if we have to calculate $\int\limits^a_b {f(x)} \, dx$
then we will divide the interval [a,b] into n subinterval of equal width.
Δx = $\frac{b-a}{n}$
So we will denote each of interval as follows
$[x_0 , x_1] , [x_1, x_2], ................[x_{n-1} , x_n]$ where $x_0 = a , x_n = b$
Then for each interval we will calculate midpoint.
So we can calculate definite integral as
$\int\limits^a_b {f(x)} \, dx = \triangle x f(y_1) + \triangle x f(y_2)+ ...............+ \triangle x f(y_n)$
where $y_1 , y_2 , ....................y_n$ are midpoint of each interval.

So in given question we need to calculate $M_6$ . So we will divide our interval in 6 equal parts.
Given interval is [1,2]
$\triangle x = \frac{2-1}{6} = \frac{1}{6}$

So we will denote 6 interval as follows
$[1 , \frac{7}{6}] , [ \frac{7}{6} , \frac{8}{6} ] , [ \frac{8}{6} , \frac{9}{6} ] , [ \frac{9}{6} , \frac{10}{6} ] , [ \frac{10}{6} , \frac{11}{6} ] , [ \frac{11}{6} , 2 ]$
Now midpoint of each interval is
$y_1 = \frac{1+ \frac{7}{6} }{2} = \frac{13}{12} = 1.084$
$y_2 = \frac{ \frac{7}{6} + \frac{8}{6} }{2} = \frac{15}{12} = 1.25$
$y_3 = \frac{ \frac{8}{6} + \frac{9}{6} }{2} = \frac{17}{12} = 1.417$
$y_4 = \frac{ \frac{9}{6} + \frac{10}{6} }{2} = \frac{19}{12} = 1.584$
$y_5 = \frac{ \frac{10}{6} + \frac{11}{6} }{2} = \frac{21}{12} = 1.75$
$y_6 = \frac{ \frac{11}{6} + \frac{12}{6} }{2} = \frac{23}{12} = 1.917$

So $M_6$ for the given function is

$M_6 = \triangle x [f(y_1) + f(y_2) + f(y_3) + f(y_4) + f(y_5) + f(y_6) ]$
                $= \frac{1}{6}[0.6452+1.7851+2.7883+3.6796+4.4769+5.4243]$
                $= \frac{1}{6} * 18.7994 = 3.1333$
So $M_6 value for function is 3.1333$