Question

If \( \int_{-\sqrt{3}}^{1} (-6x^2 + 18)\,dx = \alpha + \beta\sqrt{3} \), then the value of \( \alpha + \beta \) is equal to

• A. \(12 \)
• B. \(18 \)
• C. \(24 \)
• D. \(28 \)
• E. \(32 \)

Hint:

Always compute upper limit minus lower limit carefully with signs.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem requires the evaluation of a definite integral of a simple polynomial function. After evaluating the integral, we need to compare the result with the given form \(\alpha + \beta\sqrt{3}\) to find the values of \(\alpha\) and \(\beta\).
Step 2: Key Formula or Approach:
1. Find the antiderivative of the integrand \(-6x^2 + 18\). 2. Apply the Fundamental Theorem of Calculus: \(\int_a^b f(x)dx = F(b) - F(a)\), where F is the antiderivative of f. 3. Simplify the result and match it with the form \(\alpha + \beta\sqrt{3}\). 4. Calculate \(\alpha + \beta\).
Step 3: Detailed Explanation:
1. Find the antiderivative. \[ \int (-6x^2 + 18) dx = -6\frac{x^3}{3} + 18x = -2x^3 + 18x \] 2. Evaluate the definite integral. Let \(F(x) = -2x^3 + 18x\). We need to calculate \(F(1) - F(-\sqrt{3})\). \[ F(1) = -2(1)^3 + 18(1) = -2 + 18 = 16 \] \[ F(-\sqrt{3}) = -2(-\sqrt{3})^3 + 18(-\sqrt{3}) \] \[ = -2(-3\sqrt{3}) - 18\sqrt{3} \] \[ = 6\sqrt{3} - 18\sqrt{3} = -12\sqrt{3} \] Now, calculate the difference: \[ F(1) - F(-\sqrt{3}) = 16 - (-12\sqrt{3}) = 16 + 12\sqrt{3} \] 3. Compare and find \(\alpha\) and \(\beta\). We are given that the result is equal to \(\alpha + \beta\sqrt{3}\). \[ 16 + 12\sqrt{3} = \alpha + \beta\sqrt{3} \] By comparing the rational and irrational parts, we get:
\(\alpha = 16\)
\(\beta = 12\)
4. Calculate \(\alpha + \beta\). \[ \alpha + \beta = 16 + 12 = 28 \] Step 4: Final Answer:
The value of \(\alpha + \beta\) is 28.