MHT CET · Maths · Definite Integration
By Simpson's rule, the value of \(\int_{1}^{2} \frac{d x}{x}\) dividing the interval (1, 2) into four equal parts, is
- A \(0.6932\)
- B \(0.6753\)
- C \(0.6692\)
- D \(7.1324\)
Answer & Solution
Correct Answer
(A) \(0.6932\)
Step-by-step Solution
Detailed explanation
\(h=\frac{2-1}{4} =\frac{1}{4} \)
\( \text {Now, } x_{0} =1, x_{1}=1+\frac{1}{4}, x_{2}=1+2 \times \frac{1}{4}, \)
\( x_{3} =1+3 \times \frac{1}{4}, x_{4}=1+4 \times \frac{1}{4}\)
\( \text { ie, } x_{0} =1, x_{1}=1.25, x_{2}=1.5, x_{3} \)
\( =1.75, x_{4}=2 \)
\( \Rightarrow y_{0} =1, y_{1}=0.8, y_{2}=0.667, y_{3} \)
\( =0.571, y_{4}=0.5\)
\(\therefore\) Using Simpson's \(\frac{1}{3}\) rd rule
\(\int_{1}^{2} \frac{d x}{x}= \frac{1}{12}[(1+0.5)+4(0.8+0.571)\) \( +2(0.667)] \)
\( = \frac{1}{12}[1.5+5.484+1.334] \)
\( = \frac{1}{12}[8.318]=0.6932\)
\( \text {Now, } x_{0} =1, x_{1}=1+\frac{1}{4}, x_{2}=1+2 \times \frac{1}{4}, \)
\( x_{3} =1+3 \times \frac{1}{4}, x_{4}=1+4 \times \frac{1}{4}\)
\( \text { ie, } x_{0} =1, x_{1}=1.25, x_{2}=1.5, x_{3} \)
\( =1.75, x_{4}=2 \)
\( \Rightarrow y_{0} =1, y_{1}=0.8, y_{2}=0.667, y_{3} \)
\( =0.571, y_{4}=0.5\)
\(\therefore\) Using Simpson's \(\frac{1}{3}\) rd rule
\(\int_{1}^{2} \frac{d x}{x}= \frac{1}{12}[(1+0.5)+4(0.8+0.571)\) \( +2(0.667)] \)
\( = \frac{1}{12}[1.5+5.484+1.334] \)
\( = \frac{1}{12}[8.318]=0.6932\)
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