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

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:

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