Question Number 205315 by cherokeesay last updated on 15/Mar/24
Answered by MM42 last updated on 16/Mar/24
![[x]+[x−(1/2)]=[2x]−1 ⇒∫_0 ^([x]) ([2x]−1)dx=([2x]−1)x∣_0 ^([x]) =([2x]−1)[x] ⇒∫_0 ^([x]) (∫_0 ^([x]) ([2x]−1)[x]))dx =([2x]−1)[x]^2 ✓](https://www.tinkutara.com/question/Q205317.png)
$$\left[{x}\right]+\left[{x}−\frac{\mathrm{1}}{\mathrm{2}}\right]=\left[\mathrm{2}{x}\right]−\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\left[{x}\right]} \left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right){dx}=\left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right){x}\mid_{\mathrm{0}} ^{\left[{x}\right]} \\ $$$$=\left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right)\left[{x}\right] \\ $$$$\left.\Rightarrow\int_{\mathrm{0}} ^{\left[{x}\right]} \left(\int_{\mathrm{0}} ^{\left[{x}\right]} \left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right)\left[{x}\right]\right)\right){dx} \\ $$$$=\left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right)\left[{x}\right]^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$
Commented by cherokeesay last updated on 16/Mar/24
$${thank}\:{you}\:! \\ $$