Key Concepts and Formulas

• Definite Integration of Polynomials: The integral of a polynomial $\int (ax^n + bx^m + \dots) dx = \frac{ax^{n+1}}{n+1} + \frac{bx^{m+1}}{m+1} + \dots$ (for $n, m, \dots \neq -1$). The definite integral $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$.
• Function Evaluation: To evaluate a function $f(x)$ at a specific point, substitute that point for $x$ in the function's expression.
• Algebraic Manipulation: Simplifying expressions and finding common denominators are crucial for comparing different forms of the same mathematical quantity.
• Simpson's Rule (for reference): While not explicitly required to be known beforehand, this problem is designed such that one of the options corresponds to Simpson's 1/3 rule for approximating integrals, which is exact for quadratic (and cubic) polynomials. The rule states $\int_a^b f(x) dx \approx \frac{h}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + \dots + 4f(x_{n-1}) + f(x_n)]$, where $h = \frac{b-a}{n}$ and $x_i = a + ih$. For the interval $[0, 1]$ with $n=2$ subintervals, we have $h = \frac{1-0}{2} = \frac{1}{2}$, $x_0=0$, $x_1=1/2$, $x_2=1$. The approximation becomes $\frac{1/2}{3}[f(0) + 4f(1/2) + f(1)] = \frac{1}{6}[f(0) + 4f(1/2) + f(1)]$.

Step-by-Step Solution

Step 1: Evaluate the definite integral of $f(x)$ over $[0, 1]$. We are given the function $f(x) = a + bx + cx^2$. We need to compute $\int_0^1 f(x) dx$. $\int_0^1 (a + bx + cx^2) dx$ We integrate term by term using the power rule for integration: $\int a \, dx = ax$ $\int bx \, dx = b \frac{x^2}{2}$ $\int cx^2 \, dx = c \frac{x^3}{3}$ So, the indefinite integral is $ax + \frac{bx^2}{2} + \frac{cx^3}{3}$. Now, we evaluate this from $0$ to $1$: $\left[ ax + \frac{bx^2}{2} + \frac{cx^3}{3} \right]_0^1 = \left( a(1) + \frac{b(1)^2}{2} + \frac{c(1)^3}{3} \right) - \left( a(0) + \frac{b(0)^2}{2} + \frac{c(0)^3}{3} \right)$ $= \left( a + \frac{b}{2} + \frac{c}{3} \right) - (0)$ $= a + \frac{b}{2} + \frac{c}{3}$ To make it easier to compare with the options, we find a common denominator: $= \frac{6a}{6} + \frac{3b}{6} + \frac{2c}{6} = \frac{6a + 3b + 2c}{6}$ Let's call this result (I).

Step 2: Express the values of $f(0)$, $f(1)$, and $f(1/2)$ in terms of $a$, $b$, and $c$. The options are given in terms of $f(0)$, $f(1)$, and $f(1/2)$. We need to find these values.

• For $f(0)$: Substitute $x=0$ into $f(x) = a + bx + cx^2$. $f(0) = a + b(0) + c(0)^2 = a$
• For $f(1)$: Substitute $x=1$ into $f(x) = a + bx + cx^2$. $f(1) = a + b(1) + c(1)^2 = a + b + c$
• For $f(1/2)$: Substitute $x=1/2$ into $f(x) = a + bx + cx^2$. $f\left(\frac{1}{2}\right) = a + b\left(\frac{1}{2}\right) + c\left(\frac{1}{2}\right)^2 = a + \frac{b}{2} + \frac{c}{4}$

Step 3: Evaluate each option using the expressions found in Step 2 and compare with the result from Step 1.

Let's analyze option (A): $\frac{1}{6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$ Substitute the expressions for $f(0)$, $f(1)$, and $f(1/2)$: $\frac{1}{6}\left\{ a + (a + b + c) + 4\left(a + \frac{b}{2} + \frac{c}{4}\right) \right\}$ Simplify the expression inside the curly braces: $\frac{1}{6}\left\{ a + a + b + c + 4a + \frac{4b}{2} + \frac{4c}{4} \right\}$ $\frac{1}{6}\left\{ (a+a+4a) + (b+2b) + (c+c) \right\}$ $\frac{1}{6}\left\{ 6a + 3b + 2c \right\}$ Now, distribute the $\frac{1}{6}$: $\frac{6a}{6} + \frac{3b}{6} + \frac{2c}{6} = a + \frac{b}{2} + \frac{c}{3}$ This matches our result (I) from Step 1.

Let's quickly check other options to confirm (A) is the only correct one.

Option (B): $2\left\{ 3{f(1) + 2f\left( {{1 \over 2}} \right)} \right\}$ $2\left\{ 3(a+b+c) + 2\left(a + \frac{b}{2} + \frac{c}{4}\right) \right\}$ $2\left\{ 3a + 3b + 3c + 2a + b + \frac{c}{2} \right\}$ $2\left\{ 5a + 4b + \frac{7c}{2} \right\} = 10a + 8b + 7c \neq a + \frac{b}{2} + \frac{c}{3}$

Option (C): $\frac{1}{3}\left\{ {f(0) + f\left( {{1 \over 2}} \right)} \right\}$ $\frac{1}{3}\left\{ a + \left(a + \frac{b}{2} + \frac{c}{4}\right) \right\}$ $\frac{1}{3}\left\{ 2a + \frac{b}{2} + \frac{c}{4} \right\} = \frac{2a}{3} + \frac{b}{6} + \frac{c}{12} \neq a + \frac{b}{2} + \frac{c}{3}$

Option (D): $\frac{1}{2}\left\{ {f(1) + 3f\left( {{1 \over 2}} \right)} \right\}$ $\frac{1}{2}\left\{ (a+b+c) + 3\left(a + \frac{b}{2} + \frac{c}{4}\right) \right\}$ $\frac{1}{2}\left\{ a + b + c + 3a + \frac{3b}{2} + \frac{3c}{4} \right\}$ $\frac{1}{2}\left\{ 4a + \frac{5b}{2} + \frac{7c}{4} \right\} = 2a + \frac{5b}{4} + \frac{7c}{8} \neq a + \frac{b}{2} + \frac{c}{3}$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic simplification, especially when dealing with fractions and combining terms. A small error in arithmetic can lead to an incorrect final answer.
• Forgetting the Constant of Integration (for indefinite integrals): While this problem involves definite integrals, it's a good reminder that for indefinite integrals, the constant of integration '+ C' is essential. For definite integrals, it cancels out.
• Misinterpreting Function Notation: Ensure you correctly substitute the values into $f(x)$ and don't confuse $f(x)$ with its coefficients $a, b, c$.

Summary

The problem requires us to evaluate a definite integral of a general quadratic polynomial $f(x) = a + bx + cx^2$ over the interval $[0, 1]$. We first calculated this integral to be $a + \frac{b}{2} + \frac{c}{3}$. Then, we evaluated the function at the points $x=0$, $x=1$, and $x=1/2$, expressing these values in terms of the coefficients $a$, $b$, and $c$. Finally, we substituted these expressions into each of the given options. Option (A) yielded the exact same result as our calculated definite integral, confirming it as the correct answer. This result is also consistent with Simpson's 1/3 rule for numerical integration, which is exact for polynomials of degree up to three.

The final answer is \boxed{\frac{1}{6}\left{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right}}. This corresponds to option (A).