Question Number 86472 by Chi Mes Try last updated on 28/Mar/20
Answered by TANMAY PANACEA. last updated on 28/Mar/20
![∫_(−1) ^1 xdx−∫_(−1) ^1 [x]dx ∫_(−1) ^1 xdx−{∫_(−1) ^0 [x]dx+∫_0 ^1 [x]dx} =∣(x^2 /2)∣_(−1) ^1 −{∫_(−1) ^0 (−1)dx+∫_0 ^1 0×dx} =0−{−∣x∣_(−1) ^0 +0} =∣x∣_(−1) ^0 =(0+1)=1](https://www.tinkutara.com/question/Q86473.png)
$$\int_{−\mathrm{1}} ^{\mathrm{1}} {xdx}−\int_{−\mathrm{1}} ^{\mathrm{1}} \left[{x}\right]{dx} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} {xdx}−\left\{\int_{−\mathrm{1}} ^{\mathrm{0}} \left[{x}\right]{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \left[{x}\right]{dx}\right\} \\ $$$$=\mid\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mid_{−\mathrm{1}} ^{\mathrm{1}} −\left\{\int_{−\mathrm{1}} ^{\mathrm{0}} \left(−\mathrm{1}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{0}×{dx}\right\} \\ $$$$=\mathrm{0}−\left\{−\mid{x}\mid_{−\mathrm{1}} ^{\mathrm{0}} +\mathrm{0}\right\} \\ $$$$=\mid{x}\mid_{−\mathrm{1}} ^{\mathrm{0}} =\left(\mathrm{0}+\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$