Tuesday, 9 July 2019

definite integrals - Integration question.



The question is as follows




For any real number x, let x denote the greatest integer less than or equal to x. Let f be a real valued function defined on the interval [10,10] by




f(x)={xx  if x is odd1+xx  if x is even



Then the value of π2101010f(x) cosπx dx is




My try--




f(x) can be rewritten as



f(x)={{x}  if x is odd1{x}  if x is even



Where {} denotes the fractional part function.




The graph of f(x) will be somewhat like this, from 10 to 10.



enter image description here



So, 1010f(x) dx=10×12×2×1=10.



Then, using integration by parts, 1010f(x)2nd functioncosπx1st function dx=[cosπx10]1010+1010πsinπx10 dx      (because 1010f(x) dx=10)



But my answer does not match that given in the book.



Answer



It is easier to note that the answer is (do you see why?)

1020f(x)cos(πx)dx=1010xcos(πx)dx+1021(2x)cos(πx)dx, and then use IBP (integration by parts) on that.




EDIT: Your mistake is actually something else in your IBP. Let the integral of f(x) be denoted F(x):

1010f(x)2nd functioncosπx1st function dx=[cosπxF(x)]1010+1010πsinπxF(x) dx

When you integrate by parts, the function that isn't differentiated (that's integrated) isn't integrated definitely across the whole range but indefinitely inside the square brackets and the integral.



10uvdx=[uv]1010uvdx

Not 10uvdx=[v(10udx)]1010v(10udt)dx


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