Discussion
If \(\int_{-\sqrt{3}}^{1} (-6x^{2} + 18)\,dx = \alpha + \beta\sqrt{3}\) then the value of \(\alpha + \beta\) is equal to
Answer: D
\(\alpha + \beta = \) 28.
An antiderivative is \(F(x) = -2x^{3} + 18x\).
\[ \begin{aligned} F(1) &= -2 + 18 = 16 \\ F(-\sqrt{3}) &= -2(-3\sqrt{3}) - 18\sqrt{3} = -12\sqrt{3} \\ F(1) - F(-\sqrt{3}) &= 16 + 12\sqrt{3} \end{aligned} \]
So \(\alpha = 16\), \(\beta = 12\) and \(\alpha + \beta = 28\). A sign slip in \((-\sqrt{3})^{3} = -3\sqrt{3}\) gives \(16 + 24\sqrt{3}\) or \(16 - 12\sqrt{3}\), which lead to 40 or 4, neither offered.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development