Definite integral on fractional part function . 
Q) Let $ I_{1}=\displaystyle \int_{0}^{10^{4}}\frac{\{\sqrt{x}\}}{\sqrt{x}}dx$  and 
  $I_{2}=\displaystyle \int_{0}^{10} x\{x^{2}\}$dx .Find $I_{1},I_{2}$ ? (here {.}
  denotes fractional part of $x$)

My Attempt
For Solving $I_{1}$ we i split fractional part initially.
$$\Rightarrow 10^{4}-\int_{0}^{10^{4}}\frac{[\sqrt{x}]}{\sqrt{x}}$$ (where [.] denotes greatest integer function)   
then i took separate cases as 
$$\Rightarrow 10^{4}-\int_{0}^{1}\frac{1}{\sqrt{x}}-\int_{1}^{4}\frac{2}{\sqrt{x}}-\int_{4}^{9}\frac{3}{\sqrt{x}}-\int_{9}^{16}\frac{4}{\sqrt{x}}....$$
But i find it quite difficult to analyse terms till $10^{4}$
Please suggest me how to proceed furhter also shorter less tedious methods are more than welcome .
 A: Your method is right, except that the numerator should be 0 when the limits are from 0 to 1 and so on. The method can be made less tedious by solving for a general term n.
Basically, what we have is 
$$10^4 - \sum_{n=0}^{99}\int_{n^2}^{(n+1)^2} \frac{n}{\sqrt x} dx $$
On solving the integral, we get
$$10^4 - 2\sum_0^{99}n$$
Which gives 
$$10^4 - 99*(100)$$
$$=100$$
As for $I_2$
Substitute $t=x^2$ and hence $dt=2x dx$
Making the substitution, we get
$\frac{1}{2}\int_0^{\sqrt 10}$ {t} dt which  is periodic with T=1 and hence you get (using these properties)
$$\frac{3}{2}\int_0^1 t dt + \frac{1}{2}\int_3^{\sqrt 10} (t-3)dt$$ which can be easily evaluated.
(Similarly, for $I_1$, an alternative approach would be to take $t=\sqrt x$)
A: What immediately comes to mind for $I_1$ is to perform a substitution:  let $u = \sqrt{x}$, $du = \frac{1}{2\sqrt{x}} \, dx$, so that $$I_1 = 2\int_{u=0}^{100} \{u\} \, du = 2 \sum_{k=0}^{99} \int_{u=k}^{k+1} \{u\} \, du = 2 \sum_{k=0}^{99} \int_{u=0}^1 u \, du = 2(100)(1/2) = 100.$$
In light of this solution, can you choose an appropriate substitution for $I_2$ that facilitates its evaluation?
