MHT CET · Maths · Definite Integration
Considering four sub-intervals, the value of \(\int_{0}^{1} \frac{1}{1+x} d x\) by Trapezoidal rule, is
- A \(0.6870\)
- B \(0.6677\)
- C \(0.6977\)
- D \(0.5970\)
Answer & Solution
Correct Answer
(C) \(0.6977\)
Step-by-step Solution
Detailed explanation
| \(i\) | \(x_i\) | \(y_i=\frac{1}{1+x_i}\) |
| 0 | 0 | 1 |
| 1 | 0.25 | 0.8 |
| 2 | 0.5 | 0.67 |
| 3 | 0.75 | 0.571 |
| 4 | 1 | 0.5 |
\(\begin{aligned} \int_{0}^{1} \frac{1}{1+x} d x=& \frac{h}{2}\left[y_{0}+2\left(y_{1}+y_{2}+y_{3}\right)+y_{4}\right] \\=& \frac{1-0}{2 \times 4}[1+2(0.8+0.67\\ & \\=& 0.6977 \end{aligned}\)
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