Question

If the quadrature formula \[ \int_{-1}^{1} f(x)\,dx \;\approx\; \frac{1}{9}\Big(c_1 f(-1) + c_2 f\!\left(\tfrac{1}{2}\right) + c_3 f(1)\Big) \] is exact for all polynomials of degree less than or equal to $2$, then

• A. $c_1 + \dfrac{c_2}{4} + c_3 = 6$
• B. $c_1 + \dfrac{c_2}{3} + c_3 = 4$
• C. $c_1 + \dfrac{c_2}{2} + c_3 = 2$
• D. $c_1 + c_2 + c_3 = 5$

Hint:

“Exact for degree $\le 2$” means the rule must integrate $1$, $x$, and $x^2$ exactly—set up these three moment equations and read off the relation(s) among the weights.

Answer 1

The Correct Option is A

Step 1: Impose exactness for $f(x)=1$.
\[ \int_{-1}^{1} 1\,dx = 2 \Rightarrow \frac{1}{9}(c_1 + c_2 + c_3)=2 \;\Rightarrow\; c_1+c_2+c_3=18. \] (This relation is true but not listed among the options; keep it for checking.)
Step 2: Impose exactness for $f(x)=x$.
\[ \int_{-1}^{1} x\,dx = 0 \Rightarrow \frac{1}{9}\!\left(-c_1 + \frac{c_2}{2} + c_3\right)=0 \;\Rightarrow\; -c_1 + \frac{c_2}{2} + c_3 = 0. \]
Step 3: Impose exactness for $f(x)=x^2$.
\[ \int_{-1}^{1} x^2\,dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{2}{3}. \] RHS gives \[ \frac{1}{9}\!\left(c_1 + \frac{c_2}{4} + c_3\right)=\frac{2}{3} \;\Rightarrow\; c_1 + \frac{c_2}{4} + c_3 = 6. \] This matches option (A). (The other listed equations are inconsistent with the exactness conditions.) Final Answer: \[ \boxed{\,c_1 + \dfrac{c_2}{4} + c_3 = 6\,} \]