6
$\begingroup$

Evaluate:

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

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.

$\endgroup$
8
$\begingroup$

Let $I$ denote the given integral.

Using,

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

we get:

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

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

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

$\endgroup$
  • 1
    $\begingroup$ Nice.........., $\endgroup$ – Aqua Jul 15 '18 at 12:18
0
$\begingroup$

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

$\endgroup$
-1
$\begingroup$

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

http://www.wolframalpha.com/input/?i=graph+%7Csin(pix)%7C

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

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

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

${}{}{}$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.