Integrasi 1 - Member of EEPIS-ITS
Integrasi 1 - Member of EEPIS-ITS
Integrasi 1 - Member of EEPIS-ITS
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Contoh Metode Integral Reimann<br />
Hitung luas yang dibatasi y = x2 dan sumbu x<br />
0,<br />
1<br />
untuk range x = [ ]<br />
L =<br />
1<br />
∫<br />
0<br />
2 dx<br />
x<br />
Dengan mengambil h=0.1 maka diperoleh tabel :<br />
=<br />
x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.<br />
f(x) 0 0.01 0.04 0.04 0.09 0.16 0.25 0.36 0.49 0.64 1<br />
10<br />
L = h.<br />
∫<br />
n=<br />
=<br />
0<br />
f<br />
( x )<br />
i<br />
1<br />
0.8<br />
0.6<br />
0.4<br />
0.2<br />
0<br />
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1<br />
0.<br />
1(<br />
0 + 0.<br />
01+<br />
0.<br />
04 + 0.<br />
09 + 0.<br />
16 + 0.<br />
25 + 0.<br />
36 + 0.<br />
49 + 0.<br />
64 + 0.<br />
81+<br />
1.<br />
00)<br />
( 0.<br />
1)(<br />
3,<br />
85)<br />
= 0,<br />
385<br />
Secara kalkulus :<br />
1<br />
L = ∫ x<br />
0<br />
2<br />
1 3 1<br />
| 0<br />
dx = x<br />
3<br />
=<br />
<strong>Integrasi</strong><br />
e = 0,385-0,333<br />
= 0,052<br />
6<br />
0,<br />
3333.....<br />
x**2
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