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

Answer: (D)

17

Answer 2

To ensure continuity at $x = 1$:

$f(1^-) = 4 + a, \quad f(1^+) = b + 2.$

Setting $f(1^-) = f(1^+)$:

$4 + a = b + 2 \Rightarrow a - b = -2.$

To ensure differentiability at $x = 1$:

$f'(1^-) = 5, \quad f'(1^+) = b.$

Setting $f'(1^-) = f'(1^+)$:

$b = 5.$

Substituting $b = 5$ into $a - b = -2$:

$a = 3.$

Calculate $\int_{-2}^{2} f(x) \, dx$:

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

Evaluating each integral:

- First integral:

$\int_{-2}^{1} (x^2 + 3x + 3) \, dx = \frac{15}{2}.$

- Second integral:

$\int_{1}^{2} (5x + 2) \, dx = 17.$

The total value is:

$\int_{-2}^{2} f(x) \, dx = 17.$