Question Number 205269 by cherokeesay last updated on 14/Mar/24
Answered by som(math1967) last updated on 14/Mar/24
![∫_(−2) ^2 2f(x)dx [ ∵f(x)=f(−x)] =2∫_(−2) ^2 f(x)dx =4∫_0 ^2 f(x)dx [∵ f(x)=f(−x) ∴∫_(−2) ^2 f(x)dx=2∫_0 ^2 f(x)dx] =4×−4=−16](https://www.tinkutara.com/question/Q205270.png)
$$\:\int_{−\mathrm{2}} ^{\mathrm{2}} \mathrm{2}{f}\left({x}\right){dx}\:\:\left[\:\because{f}\left({x}\right)={f}\left(−{x}\right)\right] \\ $$$$=\mathrm{2}\int_{−\mathrm{2}} ^{\mathrm{2}} {f}\left({x}\right){dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx}\: \\ $$$$\left[\because\:{f}\left({x}\right)={f}\left(−{x}\right)\:\therefore\int_{−\mathrm{2}} ^{\mathrm{2}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx}\right] \\ $$$$=\mathrm{4}×−\mathrm{4}=−\mathrm{16} \\ $$
Commented by cherokeesay last updated on 14/Mar/24
$${thank}\:{you}. \\ $$
Answered by Sutrisno last updated on 14/Mar/24
$$\int_{−\mathrm{2}} ^{\mathrm{2}} \left({f}\left({x}\right)+{f}\left(−{x}\right)\right){dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \left({f}\left({x}\right)+{f}\left(−{x}\right)\right){dx} \\ $$$$=\mathrm{2}\left(\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left(−{x}\right){dx}\right) \\ $$$$=\mathrm{2}\left(\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx}\right) \\ $$$$=\mathrm{2}\left(−\mathrm{4}−\mathrm{4}\right) \\ $$$$=−\mathrm{16} \\ $$
Commented by cherokeesay last updated on 14/Mar/24
$${thanks}\:! \\ $$