`Let f(x)={(x^2+3x+a,xle1),(bx+2,xgt1))` is differentiable everywhere. The value of `int_(-2)^2f(x)dx` is

Options

A. `37/2`
B. `36/2`
C. `37/4`
D. `36/4`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is differentiable everywhere, especially at the point where its definition changes (i.e., at \( x = 1 \)). We will follow these steps:

### Step 1: Ensure Continuity at \( x = 1 \)

The function is defined as:
\[ f(x) = \begin{cases} x^2 + 3x + a & \text{if } x \leq 1 \\ bx + 2 & \text{if } x > 1 \end{cases} \]

For \( f(x) \) to be continuous at \( x = 1 \), we need:
\[ f(1^-) = f(1^+) \]

Calculating \( f(1^-) \):
\[ f(1^-) = 1^2 + 3(1) + a = 1 + 3 + a = 4 + a \]

Calculating \( f(1^+) \):
\[ f(1^+) = b(1) + 2 = b + 2 \]

Setting these equal for continuity:
\[ 4 + a = b + 2 \quad \text{(1)} \]

### Step 2: Ensure Differentiability at \( x = 1 \)

Next, we need to ensure that the derivatives from both sides are equal at \( x = 1 \):
\[ f'(1^-) = f'(1^+) \]

Calculating \( f'(1^-) \):
\[ f'(x) = 2x + 3 \quad \Rightarrow \quad f'(1^-) = 2(1) + 3 = 5 \]

Calculating \( f'(1^+) \):
\[ f'(x) = b \quad \Rightarrow \quad f'(1^+) = b \]

Setting these equal for differentiability:
\[ 5 = b \quad \text{(2)} \]

### Step 3: Solve for \( a \) and \( b \)

Substituting \( b = 5 \) into equation (1):
\[ 4 + a = 5 + 2 \]
\[ 4 + a = 7 \quad \Rightarrow \quad a = 3 \]

### Step 4: Calculate the Integral

Now we can express \( f(x) \) completely:
\[ f(x) = \begin{cases} x^2 + 3x + 3 & \text{if } x \leq 1 \\ 5x + 2 & \text{if } x > 1 \end{cases} \]

We need to calculate the integral:
\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{1} (x^2 + 3x + 3) \, dx + \int_{1}^{2} (5x + 2) \, dx \]

### Step 5: Evaluate the Integrals

1. **For \( \int_{-2}^{1} (x^2 + 3x + 3) \, dx \)**:
\[ \int (x^2 + 3x + 3) \, dx = \frac{x^3}{3} + \frac{3x^2}{2} + 3x \]
Evaluating from \(-2\) to \(1\):
\[ \left[ \frac{1^3}{3} + \frac{3(1^2)}{2} + 3(1) \right] - \left[ \frac{(-2)^3}{3} + \frac{3(-2)^2}{2} + 3(-2) \right] \]
\[ = \left[ \frac{1}{3} + \frac{3}{2} + 3 \right] - \left[ -\frac{8}{3} + 6 - 6 \right] \]
\[ = \left[ \frac{1}{3} + \frac{3}{2} + 3 \right] + \frac{8}{3} \]
Converting to a common denominator:
\[ = \left[ \frac{1}{3} + \frac{9}{6} + \frac{18}{6} \right] + \frac{8}{3} = \left[ \frac{1}{3} + \frac{27}{6} \right] + \frac{8}{3} \]
\[ = \frac{1 + 27}{6} + \frac{8 \cdot 2}{6} = \frac{28 + 16}{6} = \frac{44}{6} = \frac{22}{3} \]

2. **For \( \int_{1}^{2} (5x + 2) \, dx \)**:
\[ \int (5x + 2) \, dx = \frac{5x^2}{2} + 2x \]
Evaluating from \(1\) to \(2\):
\[ \left[ \frac{5(2^2)}{2} + 2(2) \right] - \left[ \frac{5(1^2)}{2} + 2(1) \right] \]
\[ = \left[ \frac{20}{2} + 4 \right] - \left[ \frac{5}{2} + 2 \right] = (10 + 4) - \left( \frac{5}{2} + 2 \right) \]
\[ = 14 - \left( \frac{5}{2} + \frac{4}{2} \right) = 14 - \frac{9}{2} = \frac{28}{2} - \frac{9}{2} = \frac{19}{2} \]

### Step 6: Combine the Results

Now, we combine the results of the two integrals:
\[ \int_{-2}^{2} f(x) \, dx = \frac{22}{3} + \frac{19}{2} \]

Finding a common denominator (which is 6):
\[ = \frac{44}{6} + \frac{57}{6} = \frac{101}{6} \]

### Final Answer

Thus, the value of \( \int_{-2}^{2} f(x) \, dx \) is:
\[ \boxed{\frac{101}{6}} \]