Key Concepts and Formulas

1. Differentiability of Piecewise Functions: For a piecewise function to be differentiable at a point $x=c$ where the definition changes, it must satisfy two conditions:

   • Continuity: The left-hand limit, right-hand limit, and the function value at $x=c$ must be equal: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
   • Equality of Derivatives: The left-hand derivative and the right-hand derivative at $x=c$ must be equal: $f'(c^-) = f'(c^+)$.

2. Definite Integration of Piecewise Functions: If a function $f(x)$ is defined piecewise, its definite integral over an interval can be split at the points where the definition changes. If $c$ is a point where the definition changes within the interval $[P, Q]$, then $\int_P^Q f(x) \, dx = \int_P^c f(x) \, dx + \int_c^Q f(x) \, dx$.

3. Basic Integration Formulas:

   • $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$)
   • $\int k \, dx = kx + C$ (where $k$ is a constant)
   • $\int_a^b g(x) \, dx = G(b) - G(a)$, where $G(x)$ is an antiderivative of $g(x)$.

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Step-by-Step Solution

Step 1: Apply the Condition for Continuity at $x=1$

For the function $f(x)$ to be differentiable at $x=1$, it must first be continuous at $x=1$. The condition for continuity is that the left-hand limit, the right-hand limit, and the function value at $x=1$ must be equal.

• The left-hand limit as $x$ approaches 1 is determined by the first piece of the function: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 3x + a)$ Substituting $x=1$: $\lim_{x \to 1^-} f(x) = (1)^2 + 3(1) + a = 1 + 3 + a = 4 + a$
• The right-hand limit as $x$ approaches 1 is determined by the second piece of the function: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (bx + 2)$ Substituting $x=1$: $\lim_{x \to 1^+} f(x) = b(1) + 2 = b + 2$
• The function value at $x=1$ is also determined by the first piece: $f(1) = (1)^2 + 3(1) + a = 4 + a$

For continuity, these must be equal: $4 + a = b + 2$ Rearranging this equation, we get our first relationship between $a$ and $b$: $a - b = -2 \quad (*)$

Step 2: Apply the Condition for Differentiability at $x=1$

For $f(x)$ to be differentiable at $x=1$, the left-hand derivative must equal the right-hand derivative at $x=1$. First, we find the derivatives of each piece of the function:

• For $x < 1$, $f(x) = x^2 + 3x + a$. The derivative is: $f'(x) = \frac{d}{dx}(x^2 + 3x + a) = 2x + 3$ The left-hand derivative at $x=1$ is: $f'(1^-) = 2(1) + 3 = 5$
• For $x > 1$, $f(x) = bx + 2$. The derivative is: $f'(x) = \frac{d}{dx}(bx + 2) = b$ The right-hand derivative at $x=1$ is: $f'(1^+) = b$

For differentiability, these must be equal: $5 = b$

Step 3: Solve for the Constants $a$ and $b$

We have found the value of $b$ from the differentiability condition: $b=5$. Now, we substitute this value of $b$ into the equation we obtained from the continuity condition ($a - b = -2$): $a - 5 = -2$ $a = -2 + 5$ $a = 3$

So, the constants are $a=3$ and $b=5$. The function $f(x)$ can now be written as: $f(x)=\left\{\begin{array}{ll}x^2+3 x+3 & , x \leq 1 \\ 5 x+2 & , x>1\end{array}\right.$

Step 4: Calculate the Definite Integral

We need to calculate $\int_{-2}^2 f(x) \, dx$. Since the definition of $f(x)$ changes at $x=1$, we split the integral at $x=1$: $\int_{-2}^2 f(x) \, dx = \int_{-2}^1 f(x) \, dx + \int_1^2 f(x) \, dx$

• First integral: For the interval $[-2, 1]$, $x \leq 1$, so we use $f(x) = x^2 + 3x + 3$: $\int_{-2}^1 (x^2 + 3x + 3) \, dx$ Find the antiderivative: $\frac{x^3}{3} + \frac{3x^2}{2} + 3x$. Evaluate from -2 to 1: $\left[ \frac{x^3}{3} + \frac{3x^2}{2} + 3x \right]_{-2}^1$ $= \left( \frac{1^3}{3} + \frac{3(1)^2}{2} + 3(1) \right) - \left( \frac{(-2)^3}{3} + \frac{3(-2)^2}{2} + 3(-2) \right)$ $= \left( \frac{1}{3} + \frac{3}{2} + 3 \right) - \left( -\frac{8}{3} + \frac{12}{2} - 6 \right)$ $= \left( \frac{2 + 9 + 18}{6} \right) - \left( -\frac{8}{3} + 6 - 6 \right)$ $= \frac{29}{6} - \left( -\frac{8}{3} \right)$ $= \frac{29}{6} + \frac{8}{3} = \frac{29}{6} + \frac{16}{6} = \frac{45}{6}$

• Second integral: For the interval $(1, 2]$, $x > 1$, so we use $f(x) = 5x + 2$: $\int_1^2 (5x + 2) \, dx$ Find the antiderivative: $\frac{5x^2}{2} + 2x$. Evaluate from 1 to 2: $\left[ \frac{5x^2}{2} + 2x \right]_1^2$ $= \left( \frac{5(2)^2}{2} + 2(2) \right) - \left( \frac{5(1)^2}{2} + 2(1) \right)$ $= \left( \frac{5(4)}{2} + 4 \right) - \left( \frac{5}{2} + 2 \right)$ $= (10 + 4) - \left( \frac{5}{2} + \frac{4}{2} \right)$ $= 14 - \frac{9}{2} = \frac{28}{2} - \frac{9}{2} = \frac{19}{2}$

Step 5: Sum the Results of the Integrals

Now, add the results from the two integrals: $\int_{-2}^2 f(x) \, dx = \frac{45}{6} + \frac{19}{2}$ To add these fractions, find a common denominator, which is 6: $\frac{45}{6} + \frac{19 \times 3}{2 \times 3} = \frac{45}{6} + \frac{57}{6}$ $= \frac{45 + 57}{6} = \frac{102}{6}$ Simplify the fraction: $\frac{102}{6} = 17$

The value of the integral is 17.

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Common Mistakes & Tips

• Forgetting Continuity: Always check for continuity before or along with the derivative condition. Differentiability implies continuity, but you need both to find the constants.
• Incorrectly Applying Derivatives: Ensure you are taking the derivative of the correct piece of the function for each side of the point $x=1$.
• Algebraic Errors in Integration: Double-check your antiderivatives and the evaluation of the definite integrals, as simple arithmetic mistakes can lead to the wrong final answer.
• Simplifying Fractions: Make sure to simplify the final fraction if possible.

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Summary

To find the values of the constants $a$ and $b$, we used the conditions for differentiability of a piecewise function at the point $x=1$. First, we ensured continuity by setting the left-hand limit equal to the right-hand limit at $x=1$, which gave us a relationship between $a$ and $b$. Second, we ensured the derivatives matched by setting the left-hand derivative equal to the right-hand derivative at $x=1$. Solving these two conditions simultaneously yielded $a=3$ and $b=5$. With the function fully defined, we then calculated the definite integral $\int_{-2}^2 f(x) \, dx$ by splitting the integral at $x=1$ and integrating each piece over its respective interval. The sum of these integrals resulted in the final answer of 17.

The final answer is \boxed{17}. This corresponds to option (C).