Question:

Simpson’s one-third rule is used to estimate the definite integral \(I=\displaystyle\int_{-1}^{1}\sqrt{1-x^{2}}\,dx\) with an interval length of \(0.5\). Which one of the following is the CORRECT estimate of \(I\) obtained using this rule?

Show Hint


For composite Simpson: \(I\approx \dfrac{h}{3}\left[f_0+f_n+4\sum f_{\text{odd}}+2\sum f_{\text{even}}\right]\) with even number of subintervals.
Recognize symmetry: values at \(\pm x\) are equal for even \(f(x)\), simplifying computations.
Updated On: Aug 26, 2025
  • \(\dfrac{1}{3}-\dfrac{1}{\sqrt{3}}\)
  • \(\dfrac{1}{3}+\dfrac{2}{\sqrt{3}}\)
  • \(\dfrac{1}{3}+\dfrac{1}{\sqrt{3}}\)
  • \(\dfrac{1}{3}-\dfrac{2}{\sqrt{3}}\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Step 1: Composite Simpson’s \(1/3\) rule with \(h=0.5\) on \([-1,1]\) uses the nodes \[ x_0=-1,\; x_1=-0.5,\; x_2=0,\; x_3=0.5,\; x_4=1. \] Let \(f(x)=\sqrt{1-x^2}\). Then \[ f(\pm1)=0,\quad f(0)=1,\quad f(\pm 0.5)=\sqrt{1-\tfrac14}=\frac{\sqrt{3}}{2}. \] Step 2: Apply the formula \[ I \approx \frac{h}{3}\Big[f(x_0)+f(x_4)+4\big(f(x_1)+f(x_3)\big)+2f(x_2)\Big] = \frac{0.5}{3}\Big[0+4\!\left(\sqrt{3}\right)+2\Big] = \frac{2\sqrt{3}+1}{3}. \] Since \(\dfrac{2\sqrt{3}}{3}=\dfrac{2}{\sqrt{3}}\), the estimate equals \(\boxed{\dfrac{1}{3}+\dfrac{2}{\sqrt{3}}}\).
Was this answer helpful?
0
0