The problem asks for the value of the expression $2a_1 + 3a_2$, based on a numerical integration formula that holds exactly for polynomials of degree up to 1.

Numerical Integration Formula Exactness

The given formula is:

$ \int_{-1}^{1} f(x)dx = a_1f \left(\frac{-1}{2}\right) + a_2f \left(\frac{1}{2}\right) $

To find the coefficients $a_1$ and $a_2$, we use the property that the formula is exact for polynomials of degree $\leq 1$. We test this with $f(x) = 1$ and $f(x) = x$.

Testing with $f(x) = 1$

Calculate the integral:

$ \int_{-1}^{1} 1 dx = [x]_{-1}^{1} = 1 - (-1) = 2 $

Apply the formula:

$ a_1 f\left(\frac{-1}{2}\right) + a_2 f\left(\frac{1}{2}\right) = a_1(1) + a_2(1) = a_1 + a_2 $

Equating both yields the first equation:

$ a_1 + a_2 = 2 \quad (\text{Equation } 1) $

Testing with $f(x) = x$

Calculate the integral:

$ \int_{-1}^{1} x dx = \left[\frac{x^2}{2}\right]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0 $

Apply the formula:

$ a_1 f\left(\frac{-1}{2}\right) + a_2 f\left(\frac{1}{2}\right) = a_1\left(\frac{-1}{2}\right) + a_2\left(\frac{1}{2}\right) = \frac{-a_1 + a_2}{2} $

Equating both gives the second equation:

$ \frac{-a_1 + a_2}{2} = 0 \implies a_1 = a_2 \quad (\text{Equation } 2) $

Solving for Coefficients $a_1$ and $a_2$

Solve the system of linear equations:

• $a_1 + a_2 = 2$
• $a_1 = a_2$

Substituting $a_1 = a_2$ into the first equation:

$ a_1 + a_1 = 2 \implies 2a_1 = 2 \implies a_1 = 1 $

Therefore, $a_2 = 1$.

Calculating $2a_1 + 3a_2$

Substitute the found values of $a_1=1$ and $a_2=1$ into the expression:

$ 2a_1 + 3a_2 = 2(1) + 3(1) = 2 + 3 = 5 $

The value of the expression $2a_1 + 3a_2$ is 5.