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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 .

Sujith Sizon
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2 Answers2

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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 $t=\sqrt x$)

Ashish Gupta
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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?

heropup
  • 135,869