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 Jun 18, 2025
-> AAI ATC exam date 2025 has been released. The exam is scheduled to be conducted on July 14.
-> AAI JE ATC recruitment 2025 application form has been released at the official website. The last date to apply for AAI ATC recruitment 2025 is May 24, 2025.
-> AAI JE ATC 2025 notification is released on 4th April 2025, along with the details of application dates, eligibility, and selection process.
-> Total number of 309 vacancies are announced for the AAI JE ATC 2025 recruitment.
-> This exam is going to be conducted for the post of Junior Executive (Air Traffic Control) in Airports Authority of India (AAI).
-> The Selection of the candidates is based on the Computer Based Test, Voice Test and Test for consumption of Psychoactive Substances.
-> The AAI JE ATC Salary 2025 will be in the pay scale of Rs 40,000-3%-1,40,000 (E-1).
-> Candidates can check the AAI JE ATC Previous Year Papers to check the difficulty level of the exam.
-> Applicants can also attend the AAI JE ATC Test Series which helps in the preparation.