Question:
If \( \int_{-\sqrt{3}}^{1} (-6x^2 + 18)\,dx = \alpha + \beta\sqrt{3} \), then the value of \( \alpha + \beta \) is equal to
If \( \int_{-\sqrt{3}}^{1} (-6x^2 + 18)\,dx = \alpha + \beta\sqrt{3} \), then the value of \( \alpha + \beta \) is equal to
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Always compute upper limit minus lower limit carefully with signs.
Updated On: Apr 21, 2026
- \(12 \)
- \(18 \)
- \(24 \)
- \(28 \)
- \(32 \)
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The Correct Option is D
Solution and Explanation
Concept:
Evaluate definite integral directly.
Step 1: Integrate.
\[ \int (-6x^2 + 18)\,dx = -2x^3 + 18x \]
Step 2: Apply limits.
\[ F(1) = -2 + 18 = 16 \] \[ F(-\sqrt{3}) = -2(-3\sqrt{3}) + 18(-\sqrt{3}) = 6\sqrt{3} - 18\sqrt{3} = -12\sqrt{3} \]
Step 3: Final value.
\[ = 16 - (-12\sqrt{3}) = 16 + 12\sqrt{3} \] \[ \alpha = 16,\quad \beta = 12 \Rightarrow \alpha + \beta = 28 \]
Step 1: Integrate.
\[ \int (-6x^2 + 18)\,dx = -2x^3 + 18x \]
Step 2: Apply limits.
\[ F(1) = -2 + 18 = 16 \] \[ F(-\sqrt{3}) = -2(-3\sqrt{3}) + 18(-\sqrt{3}) = 6\sqrt{3} - 18\sqrt{3} = -12\sqrt{3} \]
Step 3: Final value.
\[ = 16 - (-12\sqrt{3}) = 16 + 12\sqrt{3} \] \[ \alpha = 16,\quad \beta = 12 \Rightarrow \alpha + \beta = 28 \]
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