$$\int_{0}^{2008}x|\sin\pi x| dx$$

That modulus sign is causing problems. How do I handle it? I am trying integration by parts

I have even evaluated: $\int_0^1 {|\sin \pi x|}= \frac 2 \pi$. Not sure how to utilise it in the problem.

I just need help with the modulus part.


Let $I$ denote the given integral.


$\int_{a}^b f(a+b-x)dx = \int_{a}^{b} f(x)dx$

we get:

$2I = \int_0^{2008} 2008 |\sin \pi x|dx$

Using the periodicity of $|\sin \pi x|$, we obtain:

$I = \dfrac{2008^2}{\pi}$

  • 1
    $\begingroup$ Nice.........., $\endgroup$ – Aqua Jul 15 '18 at 12:18

Hint: Hence $|\sin(\pi x)|$ has a period of 1: $$\int_1^2 x|\sin(\pi x)|\,dx= \int_0^1 (1+x)|\sin(\pi x)|\,dx$$


Not sure how to bring graphs into answers, but here is a link to your function.


As you can see, the period is $1$, i.e. if

$f(x)=|\sin(\pi x)|$ then $f([0,1])=f([1,2])$.

So $\int_0^{2008}f(x) dx = 2008\int_0^1 f(x) dx$


  • $\begingroup$ there's an x in multiplication with sinpi x $\endgroup$ – Abcd Jul 4 '18 at 16:27
  • $\begingroup$ yes, you said you were unsure about how to handle this term, I thought you just needed this part, my apologies $\endgroup$ – Sheel Stueber Jul 4 '18 at 19:42

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