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)={x−⌊x⌋ if ⌊x⌋ is odd1+⌊x⌋−x if ⌊x⌋ is even
Then the value of π210∫10−10f(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.
So, ∫10−10f(x) dx=10×12×2×1=10.
Then, using integration by parts, ∫10−10f(x)⏟2nd functioncosπx⏟1st function dx=[cosπx⋅10]10−10+∫10−10πsinπx⋅10 dx (because ∫10−10f(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?)
10∗∫20f(x)cos(πx)dx=10∗∫10xcos(πx)dx+10∗∫21(2−x)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):
∫10−10f(x)⏟2nd functioncosπx⏟1st function dx=[cosπx⋅F(x)]10−10+∫10−10πsinπx⋅F(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.
∫10u′vdx=[uv]10−∫10uv′dx
Not ∫10u′vdx=[v(∫10udx)]10−∫10v′(∫10udt)dx
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