Question Number 122160 by benjo_mathlover last updated on 14/Nov/20
Answered by liberty last updated on 14/Nov/20
![∫_1 ^2 [ f(x)+1 ] dx −∫_1 ^2 [ f(t)+−1] dt = ∫_0 ^( 2) f(x)dx−∫_0 ^( 2) f(t) dt + ∫_1 ^2 dx + ∫_0 ^( 2) dt = 0 + [ x ]_1 ^2 + [ t ]_1 ^2 = 2.](Q122168.png)
$$\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\left[\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{1}\:\right]\:\mathrm{dx}\:−\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\left[\:\mathrm{f}\left(\mathrm{t}\right)+−\mathrm{1}\right]\:\mathrm{dt}\:= \\ $$$$\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}−\int_{\mathrm{0}} ^{\:\mathrm{2}} \mathrm{f}\left(\mathrm{t}\right)\:\mathrm{dt}\:+\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\mathrm{dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \mathrm{dt} \\ $$$$\:=\:\mathrm{0}\:+\:\:\left[\:\mathrm{x}\:\right]_{\mathrm{1}} ^{\mathrm{2}} \:+\:\left[\:\mathrm{t}\:\right]_{\mathrm{1}} ^{\mathrm{2}} \:=\:\mathrm{2}. \\ $$$$ \\ $$