Question Number 157016 by amin96 last updated on 18/Oct/21
$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}^{{x}} }+\frac{\mathrm{1}}{\mathrm{4}^{{x}} +\mathrm{1}}\:\: \\ $$$${find}\:\:\:\int_{\mathrm{1}} ^{\mathrm{5}} {f}\left({x}\right){dx}+\int_{−\mathrm{5}} ^{−\mathrm{1}} {f}\left({x}\right){dx} \\ $$
Answered by amin96 last updated on 18/Oct/21
![f(−x)=(2^x /(1+2^x ))+(3^x /(1+3^x ))+(4^x /(1+4^x )) f(x)+f(−x)=3 ∫_1 ^5 f(x)dx+∫_(−5) ^(−1) f(x)dx_(x=−x) ⇒ dx=−dx x[1;5] ∫_1 ^5 f(x)dx−∫_5 ^1 f(−x)dx=∫_1 ^5 (f(x)+f(−x))dx= =∫_1 ^5 3dx=3x∣_1 ^5 =15−3=12](https://www.tinkutara.com/question/Q157018.png)
$${f}\left(−{x}\right)=\frac{\mathrm{2}^{{x}} }{\mathrm{1}+\mathrm{2}^{{x}} }+\frac{\mathrm{3}^{{x}} }{\mathrm{1}+\mathrm{3}^{{x}} }+\frac{\mathrm{4}^{{x}} }{\mathrm{1}+\mathrm{4}^{{x}} } \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)=\mathrm{3} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{5}} {f}\left({x}\right){dx}+\underset{{x}=−{x}} {\underbrace{\int_{−\mathrm{5}} ^{−\mathrm{1}} {f}\left({x}\right){dx}}}\:\:\:\:\Rightarrow\:\:{dx}=−{dx}\:\:{x}\left[\mathrm{1};\mathrm{5}\right] \\ $$$$\int_{\mathrm{1}} ^{\mathrm{5}} {f}\left({x}\right){dx}−\int_{\mathrm{5}} ^{\mathrm{1}} {f}\left(−{x}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{5}} \left({f}\left({x}\right)+{f}\left(−{x}\right)\right){dx}= \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{5}} \mathrm{3}{dx}=\mathrm{3}{x}\mid_{\mathrm{1}} ^{\mathrm{5}} =\mathrm{15}−\mathrm{3}=\mathrm{12} \\ $$
Commented by amin96 last updated on 18/Oct/21
$${its}\:{true}? \\ $$
Commented by puissant last updated on 18/Oct/21
$${Good}\:{sir}.. \\ $$