Discussion
The mean value of \( f(x) = x^{2} \) over \( [0, 3] \) is
Answer: C
Divide the integral by the length of the interval:
\[ \bar{f} \;=\; \frac{1}{3}\int_{0}^{3} x^{2}dx \;=\; \frac{1}{3}\left[\frac{x^{3}}{3}\right]_{0}^{3} \;=\; \frac{1}{3}(9) \;=\; 3. \]
The mean exceeds the midpoint value \( f(1.5) = 2.25 \), as it must for a convex function, and falls well below the maximum of nine at the right-hand end
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