Question Number 222184 by hardmath last updated on 19/Jun/25
$$\mathrm{f}\left(\mathrm{2}\right)\:=\:\mathrm{8} \\ $$$$\mathrm{f}\left(\mathrm{3}\right)\:=\:\mathrm{5} \\ $$$$\int_{\mathrm{2}} ^{\:\mathrm{3}} \:\left(\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\centerdot\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$
Answered by wewji12 last updated on 20/Jun/25
![∫_2 ^( 3) (f(x)+xf′(x))dx ∫_2 ^( 3) f(x)dx+∫_2 ^( 3) xf^((1)) (x)dx (xf(x))′=f(x)+xf^((1)) (x) xf^((1)) (x)=((d )/dx)(xf(x))−f(x) ∫_( 2) ^( 3) xf^((1)) (x)dx=[xf(x)]_(x=2) ^(x=3) −∫_2 ^( 3) f(x)dx ∴ ∫_2 ^( 3) (f(x)+xf^((1)) (x))dx= ∫_2 ^( 3) f(x)dx+ [xf(x)]_(x=2) ^(x=3) −∫_2 ^( 3) f(x)dx ∫_2 ^( 3) f(x)dx+[xf(x)]_(x=2) ^(x=3) −∫_2 ^( 3) f(x)dx=[xf(x)]_(x=2) ^(x=3) 3f(3)−2f(2)=15−16=−1 ∴ ∫_2 ^( 3) {f(x)−xf^((1)) (x)}dx=−1](https://www.tinkutara.com/question/Q222186.png)
$$\int_{\mathrm{2}} ^{\:\mathrm{3}} \left({f}\left({x}\right)+{xf}'\left({x}\right)\right)\mathrm{d}{x} \\ $$$$\int_{\mathrm{2}} ^{\:\mathrm{3}} \:{f}\left({x}\right)\mathrm{d}{x}+\int_{\mathrm{2}} ^{\:\mathrm{3}} \:{xf}^{\left(\mathrm{1}\right)} \left({x}\right)\mathrm{d}{x} \\ $$$$\left({xf}\left({x}\right)\right)'={f}\left({x}\right)+{xf}^{\left(\mathrm{1}\right)} \left({x}\right) \\ $$$${xf}^{\left(\mathrm{1}\right)} \left({x}\right)=\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left({xf}\left({x}\right)\right)−{f}\left({x}\right) \\ $$$$\int_{\:\mathrm{2}} ^{\:\mathrm{3}} \:{xf}^{\left(\mathrm{1}\right)} \left({x}\right)\mathrm{d}{x}=\left[{xf}\left({x}\right)\right]_{{x}=\mathrm{2}} ^{{x}=\mathrm{3}} −\int_{\mathrm{2}} ^{\:\mathrm{3}} \:{f}\left({x}\right)\mathrm{d}{x} \\ $$$$\therefore\:\int_{\mathrm{2}} ^{\:\mathrm{3}} \:\left({f}\left({x}\right)+{xf}^{\left(\mathrm{1}\right)} \left({x}\right)\right)\mathrm{d}{x}= \\ $$$$\int_{\mathrm{2}} ^{\:\mathrm{3}} \:{f}\left({x}\right)\mathrm{d}{x}+\:\left[{xf}\left({x}\right)\right]_{{x}=\mathrm{2}} ^{{x}=\mathrm{3}} −\int_{\mathrm{2}} ^{\:\mathrm{3}} \:{f}\left({x}\right)\mathrm{d}{x} \\ $$$$\cancel{\int_{\mathrm{2}} ^{\:\mathrm{3}} \:{f}\left({x}\right)\mathrm{d}{x}}+\left[{xf}\left({x}\right)\right]_{{x}=\mathrm{2}} ^{{x}=\mathrm{3}} −\cancel{\int_{\mathrm{2}} ^{\:\mathrm{3}} \:{f}\left({x}\right)\mathrm{d}{x}}=\left[{xf}\left({x}\right)\right]_{{x}=\mathrm{2}} ^{{x}=\mathrm{3}} \\ $$$$\mathrm{3}{f}\left(\mathrm{3}\right)−\mathrm{2}{f}\left(\mathrm{2}\right)=\mathrm{15}−\mathrm{16}=−\mathrm{1} \\ $$$$\therefore\:\int_{\mathrm{2}} ^{\:\mathrm{3}} \:\left\{{f}\left({x}\right)−{xf}^{\left(\mathrm{1}\right)} \left({x}\right)\right\}\mathrm{d}{x}=−\mathrm{1} \\ $$
Answered by mr W last updated on 20/Jun/25
![∫_2 ^3 (f(x)+f′(x)x)dx =∫_2 ^3 d(xf(x)) =[xf(x)]_2 ^3 =3f(3)−2f(2)=3×5−2×8=−1](https://www.tinkutara.com/question/Q222188.png)
$$\int_{\mathrm{2}} ^{\mathrm{3}} \left({f}\left({x}\right)+{f}'\left({x}\right){x}\right){dx} \\ $$$$=\int_{\mathrm{2}} ^{\mathrm{3}} {d}\left({xf}\left({x}\right)\right) \\ $$$$=\left[{xf}\left({x}\right)\right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$=\mathrm{3}{f}\left(\mathrm{3}\right)−\mathrm{2}{f}\left(\mathrm{2}\right)=\mathrm{3}×\mathrm{5}−\mathrm{2}×\mathrm{8}=−\mathrm{1} \\ $$
Commented by hardmath last updated on 21/Jun/25
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$