Question

For what value(s) of the constants \(a\), \(b\) and \(c\) is the quadrature formula \[\int_{-1}^1 f(x)\,dx \approx af(-1)+bf(0)+cf(1)\] exact for polynomials of degree up to 3?

• A. a = 1/3, b = 4/3 and c = 1/3
• B. a = 4/3, b = 1/3 and c = 1/3
• C. a = 1/3, b = 1/3 and c = 4/3
• D. a = 1/3, b = 4/3 and c = -1/3

Hint:

Match the formula against Simpson's 1/3 rule on [-1, 1], or force exactness for 1, x and x^2.

Answer 1

The Correct Option is A

Step 1: Impose exactness on the monomials $1, x, x^2$; by symmetry of the interval $[-1,1]$, exactness for $x^3$ then follows automatically since both sides vanish for odd powers once $a=c$.

Step 2: For $f(x)=1$:\[\int_{-1}^1 1\,dx = 2 = a(1)+b(1)+c(1) \ \Rightarrow\ a+b+c=2\]

Step 3: For $f(x)=x$:\[\int_{-1}^1 x\,dx = 0 = a(-1)+b(0)+c(1) \ \Rightarrow\ -a+c=0 \ \Rightarrow\ a=c\]

Step 4: For $f(x)=x^2$:\[\int_{-1}^1 x^2\,dx=\frac{2}{3} = a(1)+b(0)+c(1) \ \Rightarrow\ a+c=\frac{2}{3}\]

Step 5: Since $a=c$, Step 4 gives $2a=\frac{2}{3}\Rightarrow a=c=\frac{1}{3}$. Substituting into Step 2, $b=2-a-c=2-\frac{1}{3}-\frac{1}{3}=\frac{4}{3}$.

Step 6: Check $f(x)=x^3$ (an odd function): both $\int_{-1}^1x^3\,dx=0$ and $a(-1)+b(0)+c(1)=-a+c=0$ hold automatically since $a=c$. So the formula is exact up to degree $3$ with these constants.

\[\boxed{a=\frac{1}{3},\ b=\frac{4}{3},\ c=\frac{1}{3}}\]