MHT CET · Maths · Definite Integration
By the application of Simpson's one-third rule for numerical integration, with two subintervals, the value of \(\int_{0}^{1} \frac{d x}{1+x}\) is
- A \(\frac{17}{36}\)
- B \(\frac{17}{25}\)
- C \(\frac{25}{36}\)
- D \(\frac{17}{24}\)
Answer & Solution
Correct Answer
(C) \(\frac{25}{36}\)
Step-by-step Solution
Detailed explanation
Since, the given integration is divided into two subintervals.
ie,
\(
h=\frac{1-0}{2}=\frac{1}{2}
\)
\(
\therefore \int_{0}^{1} \frac{1}{1+x} d x=\frac{h}{3}\left[\left(y_{0}+y_{2}\right)+4\left(y_{1}\right)\right]
\)
At
\(x=0, y =1 \)
\( x =\frac{1}{2}, y_{1}=\frac{2}{3}\)
and
\(
x=1, y_{2}=\frac{1}{2}
\)
\(\therefore \int_{0}^{1} \frac{1}{1+x} d x =\frac{1}{2 \cdot 3}\left[\left(1+\frac{1}{2}\right)+4\left(\frac{2}{3}\right)\right] \)
\( =\frac{1}{6}\left[\frac{3}{2}+\frac{8}{3}\right]=\frac{25}{36}\)
ie,
\(
h=\frac{1-0}{2}=\frac{1}{2}
\)
\(
\therefore \int_{0}^{1} \frac{1}{1+x} d x=\frac{h}{3}\left[\left(y_{0}+y_{2}\right)+4\left(y_{1}\right)\right]
\)
At
\(x=0, y =1 \)
\( x =\frac{1}{2}, y_{1}=\frac{2}{3}\)
and
\(
x=1, y_{2}=\frac{1}{2}
\)
\(\therefore \int_{0}^{1} \frac{1}{1+x} d x =\frac{1}{2 \cdot 3}\left[\left(1+\frac{1}{2}\right)+4\left(\frac{2}{3}\right)\right] \)
\( =\frac{1}{6}\left[\frac{3}{2}+\frac{8}{3}\right]=\frac{25}{36}\)
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