Quadrature Formula Exactness

A quadrature formula is considered exact for polynomials of a certain degree if it yields the precise value of the integral for any polynomial of that degree or less. The given quadrature formula is:

\(\int_{-1}^{1} f(x)dx \approx \frac{1}{9} \left(c_1f(-1) + c_2f\left(\frac{1}{2}\right) + c_3f(1)\right)\)

We are given that this formula is exact for all polynomials of degree less than or equal to 2. This means it must hold true for the basis polynomials: $f(x) = 1$, $f(x) = x$, and $f(x) = x^2$. We test these polynomials to find the relationship between the coefficients $c_1$, $c_2$, and $c_3$.

Testing Polynomial Exactness

• Test with $f(x) = x^2$:

  The exact integral is:

  \(\int_{-1}^{1} x^2 dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3}\)

  Applying the quadrature formula:

  \(\frac{1}{9} \left(c_1f(-1) + c_2f\left(\frac{1}{2}\right) + c_3f(1)\right) = \frac{1}{9} \left(c_1(-1)^2 + c_2\left(\frac{1}{2}\right)^2 + c_3(1)^2\right)\)

  \(= \frac{1}{9} \left(c_1(1) + c_2\left(\frac{1}{4}\right) + c_3(1)\right) = \frac{1}{9} \left(c_1 + \frac{c_2}{4} + c_3\right)\)

  For exactness, the results must be equal:

  \(\frac{1}{9} \left(c_1 + \frac{c_2}{4} + c_3\right) = \frac{2}{3}\)

  Multiplying both sides by 9 gives:

  \(c_1 + \frac{c_2}{4} + c_3 = 9 \times \frac{2}{3}\)

  \(c_1 + \frac{c_2}{4} + c_3 = 6\)

This derived relationship, $c_1 + \frac{c_2}{4} + c_3 = 6$, matches one of the options provided.