integrate with limits 3..and 1 [x-1] +[x-2] + [x-3]dx
Answer (1)
15th Jan, 2021
Hello Candidate,
X is an integer, so the terms x-1, x-3, x-2, are all Integers.
[x] refers to the greatest number Integer of x, and is referred as [x]= x - {x}, and {x} refers to the fractional part of x and will be zero in this question, since x is a pure Integer.
So, (x-1 + x-2 + x-3) dx= (3x- 6) dx= 3/2 x^2 - 6x +Constant within limits 1 and 3 gives the final answer as 3/2× (9-1) + 6×(3-1)= 0.
Hope that it helps!! Have a Good Day.
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