1 1 vote If the trapezoidal method is used to evaluate the integral obtained $\int_{0}^{1} x^2dx$, then the value obtained is always > (1/3) is always < (1/3) is always = (1/3) may be greater or lesser than (1/3) Numerical Methods gateit-2005 numerical-methods trapezoidal-rule normal + – Ishrat Jahan 2.6k views answer comment Share Follow Print See 1 comment 1 1 comment reply P0535_Yedidyah_Sagar commented Sep 16, 2025 reply Follow flag Out of syllabus 1 1 replyShare Please log in or register to add a comment.
0 0 votes Answer is (A) The approximate value calculated by the trapezoidal method is always greater than the actual value. This can be supported by the argument that value of ERROR= EXACT VALUE- APPROXIMATED VALUE comes out to be NEGATIVE which shows that approximated value is greater. Sandeep_Uniyal answered Jan 17, 2015 Sandeep_Uniyal comment Share Follow See all 3 Comments 3 3 Comments reply Arjun commented Jan 18, 2015 reply Follow flag Not for all functions rt? http://tutorial.math.lamar.edu/Classes/CalcII/ApproximatingDefIntegrals.aspx 1 1 replyShare Sandeep_Uniyal commented Jan 18, 2015 reply Follow flag But the given function is strictly decreasing from 0 1o 1 . can we use this term :"non-monotonically" ?? A function is either "monotonically increasing or strictly increasing or None " It overestimates the value for all the non-linear curves http://en.wikipedia.org/wiki/Trapezoidal_rule 0 0 replyShare Arjun commented Jan 18, 2015 reply Follow flag yes. I meant that the answer is specific to this question and not applicable for all functions. I don't know exactly which all functions it can be true but I guess it won't be true for any functions which is not monotonically increasing or decreasing. In the wiki page it is given- overestimate for all concave curves and underestimate for all convex curves. 2 2 replyShare Please log in or register to add a comment.
