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Using centro id mean of maxima and center of sums controlled apply defuzzification.
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[1] Centro-id method.

Computation of $x^x$

Area segment No Area (A) $\bar x$ $A( \bar x )$
1 $\frac{1}{2} \times 0.3 \times 1 = 0.15$ $\frac{0 + 1 + 1}{3} = 0.67$ 0.1005
2 $2.5 \times 3 = 0.78$ (1 + 3.6) 12 = 2.3 1.794
3 $0.4 \times 0.3 = 0.12$ (3.6 + 4) 12 = 3.8 0.456
4 $\frac{1}{2} \times 0.4 \times 0.2 = 0.04$ 3.8667 0.1547
5 $1.5 \times 0.5 = 0.75$ 4.75 3.56
6 $0.5 \times 0.5 = 0.25$ 5.75 1.4375
7 $\frac{1}{2} \times 0.5 \times 0.5 = 0.125$ 5.833 0.729
8 $1 \times 1 = 1$ 6.5 6.5
9 $\frac{1}{2} \times 1 \times 1 = 0.5$ 7.33 3.665

$x^* = \frac{\sum a ( \bar X)}{\sum A} = \frac{18.3967}{3.7} = 4.9$

[b] Weighted Average : (valid for symmetric mean)

$x^* = \frac{2.5 * 0.3 + 5 \times 0.5 + 6.5 * 1}{0.3 + 0.5 + 1}$

= 5.41

[c] Centre of sums:

Area of $A_1$ = $\frac{1}{2} \times 0.3 \times 1 + 3 \times 0.3 + \frac{1}{2} \times 1 \times 0.3$

= 0.15 + 0.9 + 0.15

= 1.2

Area of $A_2$ = $\frac{1}{2} \times 1 \times 0.5 + 2 \times 0.5 + \frac{1}{2} \times 1 + 0.5$

= 0.25 + 1.0 + 0.25

= 1.5

Area of $A_3$ = $\frac{1}{2} \times 1 \times 1 + 1 \times 1 + \frac{1}{2} \times 1 \times 1$

$= \frac{1}{2} + 1 + \frac{1}{2} = 2$

$x^x = \frac{1.2 * 2.5 + 1.5 * 5 + 2 * 6.5}{1.2 + 1.5 + 2}$

$= \frac{23.5}{4.7} = 5$

[d] First of maxima (last of maxima):

$A_3$ is having the largest area.

First of maxima = 6 Last of maxima 7

$x^* = \frac{6 + 7}{2} = 6.5$