Question
Download Solution PDFचार उप-अंतरालों को ध्यान में रखते हुए, समलम्बाकार नियम द्वारा \(\mathop \smallint \limits_0^1 \frac{1}{{1 + x}}dx\) का मान है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
समलम्बाकार नियम कहता है कि:
\(\mathop \smallint \limits_{{x_0}}^{{x_0} + nh} f\left( x \right)\;dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + \ldots + {y_{n - 1}}} \right)} \right]\)
गणना:
\(\mathop \smallint \limits_0^1 \frac{1}{{1 + x}}dx\)
अंतराल की कुल संख्या = 4
|
x |
0 |
0.25 |
0.5 |
0.75 |
1 |
|
f(x) |
y0 = 1 |
y1 = 0.8 |
y2 = 0.66 |
y3 = 0.57 |
y4 = 0.5 |
\(h = \frac{1}{4} = 0.25\)
\(\mathop \smallint \limits_0^1 \frac{1}{{1 + x}}dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_4}} \right) + 2\left( {{y_1} + {y_2} + {y_3}} \right)} \right]\)
\(= \frac{1}{8}\left[ {\left( {1 + 0.5} \right) + 2\left( {0.8 + 0.66 + 0.57} \right)} \right]\)
\(\mathop \smallint \limits_0^1 \frac{1}{{1 + x}}dx = 0.6950 \)
Last updated on May 26, 2025
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