Question

[Integrate {sin(sqr. x)/sqr.] x dx
Upper limit pi^2
Lower limit 0+

Answer 1

\[\int\limits_{0}^{\pi^2}\frac{\sin(\sqrt{x})}{\sqrt{x}}dx\]
is that the question?

Answer 2

yes, only the 0 has a + sign on top

Answer 3

Yeah, I dont think that really matters...lol
So, start by using substitution:
Let u= sqrt (x)
du = 1/(2 sqrt (x))
and you will see:
\[2\int \sin(u) du\]
\[-2\cos(u)\]
\[-2\cos(\sqrt x)\]
evaluate from your limits.

Answer 4

should be 4.