Question

What is the approximate value of the integral \(\int_0^1(x+x^2)\,dx\), when Simpson's 1/3-rd rule is applied with 2 sub-intervals?

• A. 0.83
• B. 0.75
• C. 0.67
• D. 0.58

Hint:

Use nodes at x = 0, 0.5, 1 and apply the weighted sum h/3[f0+4f1+f2].

Answer 1

The Correct Option is A

Step 1: With 2 sub-intervals over $[0,1]$, the step size is\[h=\frac{1-0}{2}=0.5\]giving the nodes $x_0=0$, $x_1=0.5$, $x_2=1$.

Step 2: Evaluate $f(x)=x+x^2$ at the nodes.\[f(x_0)=f(0)=0,\quad f(x_1)=f(0.5)=0.5+0.25=0.75,\quad f(x_2)=f(1)=1+1=2\]

Step 3: Apply Simpson's 1/3 rule,\[\int_0^1 f(x)\,dx\approx \frac{h}{3}\left[f(x_0)+4f(x_1)+f(x_2)\right]=\frac{0.5}{3}\left[0+4(0.75)+2\right]\]\[=\frac{0.5}{3}\left[0+3+2\right]=\frac{0.5}{3}(5)=\frac{2.5}{3}\approx 0.83\]

\[\boxed{\int_0^1(x+x^2)\,dx\approx 0.83}\]