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
- \( \frac{15}{6} \)
- \( \frac{19}{6} \)
- 21
- 17
The Correct Option is D
Approach Solution - 1
We are asked to find the value of the definite integral \( \int_{-2}^{2} f(x) \, dx \) for the given piecewise function, which is stated to be differentiable over \( \mathbb{R} \). This requires us to first find the values of the constants \( a \) and \( b \) by using the conditions for differentiability.
Concept Used:
1. Continuity: For a function to be differentiable at a point, it must first be continuous at that point. For the given piecewise function, continuity at the boundary point \( x=1 \) implies that the left-hand limit (LHL) must equal the right-hand limit (RHL).
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]
2. Differentiability: For the function to be differentiable at \( x=1 \), the left-hand derivative (LHD) must equal the right-hand derivative (RHD) at that point.
\[ f'_{-}(1) = f'_{+}(1) \]
3. Integration of a Piecewise Function: To evaluate the definite integral of a piecewise function over an interval, we must split the integral at the points where the function's definition changes.
\[ \int_{c}^{d} f(x) \, dx = \int_{c}^{k} f_1(x) \, dx + \int_{k}^{d} f_2(x) \, dx \] where the function definition changes at \( x=k \).
Step-by-Step Solution:
Step 1: Apply the condition of continuity at \( x = 1 \).
The left-hand limit is:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 3x + a) = 1^2 + 3(1) + a = 4 + a \]
The right-hand limit is:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (bx + 2) = b(1) + 2 = b + 2 \]
For continuity, we must have LHL = RHL:
\[ 4 + a = b + 2 \implies b - a = 2 \quad \text{ (Equation 1)} \]
Step 2: Apply the condition of differentiability at \( x = 1 \).
First, we find the derivatives for each piece of the function.
For \( x < 1 \), \( f'(x) = \frac{d}{dx}(x^2 + 3x + a) = 2x + 3 \). The left-hand derivative at \( x=1 \) is:
\[ f'_{-}(1) = 2(1) + 3 = 5 \]
For \( x > 1 \), \( f'(x) = \frac{d}{dx}(bx + 2) = b \). The right-hand derivative at \( x=1 \) is:
\[ f'_{+}(1) = b \]
For differentiability, we must have LHD = RHD:
\[ b = 5 \]
Step 3: Solve for the constants \( a \) and \( b \).
We found \( b = 5 \). Substituting this value into Equation 1:
\[ 5 - a = 2 \implies a = 3 \]
So, the function is:
\[ f(x) = \begin{cases} x^2 + 3x + 3, & x \leq 1 \\ 5x + 2, & x>1 \end{cases} \]
Step 4: Evaluate the definite integral \( \int_{-2}^{2} f(x) \, dx \).
Since the function definition changes at \( x = 1 \), we split the integral at this point.
\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{1} (x^2 + 3x + 3) \, dx + \int_{1}^{2} (5x + 2) \, dx \]
Step 5: Calculate each integral separately.
For the first integral:
\[ \int_{-2}^{1} (x^2 + 3x + 3) \, dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} + 3x \right]_{-2}^{1} \] \[ = \left( \frac{1^3}{3} + \frac{3(1)^2}{2} + 3(1) \right) - \left( \frac{(-2)^3}{3} + \frac{3(-2)^2}{2} + 3(-2) \right) \] \[ = \left( \frac{1}{3} + \frac{3}{2} + 3 \right) - \left( -\frac{8}{3} + \frac{12}{2} - 6 \right) \] \[ = \left( \frac{2+9+18}{6} \right) - \left( -\frac{8}{3} + 6 - 6 \right) = \frac{29}{6} - \left( -\frac{8}{3} \right) = \frac{29}{6} + \frac{16}{6} = \frac{45}{6} = \frac{15}{2} \]
For the second integral:
\[ \int_{1}^{2} (5x + 2) \, dx = \left[ \frac{5x^2}{2} + 2x \right]_{1}^{2} \] \[ = \left( \frac{5(2)^2}{2} + 2(2) \right) - \left( \frac{5(1)^2}{2} + 2(1) \right) \] \[ = (10 + 4) - \left( \frac{5}{2} + 2 \right) = 14 - \frac{9}{2} = \frac{28 - 9}{2} = \frac{19}{2} \]
Final Computation & Result:
Add the results of the two integrals to get the final answer.
\[ \int_{-2}^{2} f(x) \, dx = \frac{15}{2} + \frac{19}{2} = \frac{34}{2} = 17 \]
The value of the integral is 17.
Approach Solution -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:
\(\int_{-2}^{2} f(x) \, dx = 17.\)
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