Question Number 199934 by cortano12 last updated on 11/Nov/23
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Answered by mr W last updated on 11/Nov/23
![∫_0 ^3 f(x)dx=a, say ∫_0 ^3 ((f(3x))/3)dx=a (1/9)∫_0 ^3 f(3x)d(3x)=a (1/9)∫_0 ^9 f(t)dt=a (1/9)∫_0 ^9 ((f(3t))/3)dt=a (1/(9×9))∫_0 ^9 f(3t)d(3t)=a (1/(9×9))∫_0 ^(27) f(u)du=a (1/(9×9))[∫_0 ^3 f(x)dx+∫_3 ^(27) f(x)dx]=a (1/(9×9))[a+10]=a 10=80a ⇒a=(1/8)](Q199937.png)
$$\int_{\mathrm{0}} ^{\mathrm{3}} {f}\left({x}\right){dx}={a},\:{say} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} \frac{{f}\left(\mathrm{3}{x}\right)}{\mathrm{3}}{dx}={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{3}} {f}\left(\mathrm{3}{x}\right){d}\left(\mathrm{3}{x}\right)={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{9}} {f}\left({t}\right){dt}={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{9}} \frac{{f}\left(\mathrm{3}{t}\right)}{\mathrm{3}}{dt}={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}×\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{9}} {f}\left(\mathrm{3}{t}\right){d}\left(\mathrm{3}{t}\right)={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}×\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{27}} {f}\left({u}\right){du}={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}×\mathrm{9}}\left[\int_{\mathrm{0}} ^{\mathrm{3}} {f}\left({x}\right){dx}+\int_{\mathrm{3}} ^{\mathrm{27}} {f}\left({x}\right){dx}\right]={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}×\mathrm{9}}\left[{a}+\mathrm{10}\right]={a} \\ $$$$\mathrm{10}=\mathrm{80}{a} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by cortano12 last updated on 11/Nov/23
$$\:\:\underbrace{\Subset} \\ $$