Question

Let \( a \) and \( b \) be real constants such that the function \( f \) defined by \[f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 \\ bx + 2, & x>1 \end{cases}\]be differentiable on \( \mathbb{R} \). Then, the value of \( \int_{-2}^{2} f(x) \, dx \) equals

• A. \( \frac{15}{6} \)
• B. \( \frac{19}{6} \)
• C. 21
• D. 17

Answer 1

The Correct Option is D

The objective is to compute the definite integral \( \int_{-2}^{2} f(x) \, dx \) of a given differentiable piecewise function. This necessitates determining the unknown constants \( a \) and \( b \) using differentiability criteria.

Key Concepts:

1. Continuity: A prerequisite for differentiability at a point is continuity. For the piecewise function at \( x=1 \), continuity requires the left-hand limit (LHL) to equal the right-hand limit (RHL).

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]

2. Differentiability: For differentiability at \( x=1 \), the left-hand derivative (LHD) must equal the right-hand derivative (RHD).

\[ f'_{-}(1) = f'_{+}(1) \]

3. Piecewise Integral Evaluation: The definite integral of a piecewise function over an interval is computed by segmenting the integral at each point where the function definition changes.

\[ \int_{c}^{d} f(x) \, dx = \int_{c}^{k} f_1(x) \, dx + \int_{k}^{d} f_2(x) \, dx \] where \( x=k \) is a point of definition change.

Solution Steps:

Step 1: Enforce continuity at \( x = 1 \).

Left-hand limit:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 3x + a) = 1 + 3 + a = 4 + a \]

Right-hand limit:

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (bx + 2) = b + 2 \]

Equating limits for continuity:

\[ 4 + a = b + 2 \implies b - a = 2 \quad \text{(Equation 1)} \]

Step 2: Apply the differentiability condition at \( x = 1 \).

Derivatives of the function pieces:

For \( x<1 \), \( f'(x) = 2x + 3 \). Left-hand derivative at \( x=1 \):

\[ f'_{-}(1) = 2(1) + 3 = 5 \]

For \( x>1 \), \( f'(x) = b \). Right-hand derivative at \( x=1 \):

\[ f'_{+}(1) = b \]

Equating derivatives for differentiability:

\[ b = 5 \]

Step 3: Determine constants \( a \) and \( b \).

With \( b = 5 \), substitute into Equation 1:

\[ 5 - a = 2 \implies a = 3 \]

The function is thus:

\[ f(x) = \begin{cases} x^2 + 3x + 3, & x \leq 1 \\ 5x + 2, & x>1 \end{cases} \]

Step 4: Compute the definite integral \( \int_{-2}^{2} f(x) \, dx \).

Split the integral at \( x = 1 \):

\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{1} (x^2 + 3x + 3) \, dx + \int_{1}^{2} (5x + 2) \, dx \]

Step 5: Evaluate each integral.

First integral:

\[ \int_{-2}^{1} (x^2 + 3x + 3) \, dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} + 3x \right]_{-2}^{1} \] \[ = \left( \frac{1}{3} + \frac{3}{2} + 3 \right) - \left( -\frac{8}{3} + 6 - 6 \right) \] \[ = \frac{29}{6} + \frac{8}{3} = \frac{29+16}{6} = \frac{45}{6} = \frac{15}{2} \]

Second integral:

\[ \int_{1}^{2} (5x + 2) \, dx = \left[ \frac{5x^2}{2} + 2x \right]_{1}^{2} \] \[ = \left( \frac{5(4)}{2} + 4 \right) - \left( \frac{5}{2} + 2 \right) \] \[ = (10 + 4) - \frac{9}{2} = 14 - \frac{9}{2} = \frac{28 - 9}{2} = \frac{19}{2} \]

Final Calculation & Result:

Sum the results of the two integrals:

\[ \int_{-2}^{2} f(x) \, dx = \frac{15}{2} + \frac{19}{2} = \frac{34}{2} = 17 \]

The value of the integral is 17.