Question Number 85888 by M±th+et£s last updated on 25/Mar/20
![i−∫_0 ^5 (x+[2x])^([(x/3)]) dx ii−∫_0 ^3 (z−{z})^([z]) dz](Q85888.png)
$${i}−\int_{\mathrm{0}} ^{\mathrm{5}} \left({x}+\left[\mathrm{2}{x}\right]\right)^{\left[\frac{{x}}{\mathrm{3}}\right]} {dx} \\ $$$$ \\ $$$${ii}−\int_{\mathrm{0}} ^{\mathrm{3}} \left({z}−\left\{{z}\right\}\right)^{\left[{z}\right]} \:{dz} \\ $$
Commented by abdomathmax last updated on 25/Mar/20
![let I =∫_0 ^5 (x+[2x])^([(x/3)]) dx changement (x/3)=t give I =3∫_0 ^(5/3) (3t +[6t])^([t]) dt =3( ∫_0 ^1 (3t+[6t])^([t]) dt +∫_1 ^(5/3) (3t+[6t])^([t]) dt) =3{ ∫_0 ^1 (3t+[6t])^0 dt +∫_1 ^(5/3) (3t+[6t])dt} =3 +9∫_1 ^(5/3) t dt +3∫_1 ^(5/3) [6t]dt we have ∫_1 ^(5/3) t dt =[(t^2 /2)]_1 ^(5/3) =((25)/(18))−(1/2) ∫_1 ^(5/3) [6t]dt =_(6t=u) (1/6) ∫_6 ^(10) [u] du =(1/6)Σ_(k=6) ^9 ∫_k ^(k+1) k du =(1/6)Σ_(k=6) ^9 k =(1/6)(6+7+8+9) =((13+17)/6) =((30)/6)=5 ⇒ I = 3 +9(((25)/(18))−(1/2)) +15 =18+((25)/2)−(9/2) =18 +8 =26](Q85894.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{5}} \left({x}+\left[\mathrm{2}{x}\right]\right)^{\left[\frac{{x}}{\mathrm{3}}\right]} \:{dx}\:\:{changement}\:\frac{{x}}{\mathrm{3}}={t}\:{give} \\ $$$${I}\:=\mathrm{3}\int_{\mathrm{0}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}{t}\:+\left[\mathrm{6}{t}\right]\right)^{\left[{t}\right]} {dt} \\ $$$$=\mathrm{3}\left(\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3}{t}+\left[\mathrm{6}{t}\right]\right)^{\left[{t}\right]} \:{dt}\:+\int_{\mathrm{1}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}{t}+\left[\mathrm{6}{t}\right]\right)^{\left[{t}\right]} {dt}\right) \\ $$$$=\mathrm{3}\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3}{t}+\left[\mathrm{6}{t}\right]\right)^{\mathrm{0}} \:{dt}\:+\int_{\mathrm{1}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}{t}+\left[\mathrm{6}{t}\right]\right){dt}\right\} \\ $$$$=\mathrm{3}\:+\mathrm{9}\int_{\mathrm{1}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \:{t}\:{dt}\:+\mathrm{3}\int_{\mathrm{1}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \left[\mathrm{6}{t}\right]{dt}\:\:{we}\:{have} \\ $$$$\int_{\mathrm{1}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \:{t}\:{dt}\:=\left[\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \:=\frac{\mathrm{25}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\frac{\mathrm{5}}{\mathrm{3}}} \left[\mathrm{6}{t}\right]{dt}\:=_{\mathrm{6}{t}={u}} \:\:\:\frac{\mathrm{1}}{\mathrm{6}}\:\int_{\mathrm{6}} ^{\mathrm{10}} \left[{u}\right]\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\sum_{{k}=\mathrm{6}} ^{\mathrm{9}} \:\int_{{k}} ^{{k}+\mathrm{1}} {k}\:{du}\:=\frac{\mathrm{1}}{\mathrm{6}}\sum_{{k}=\mathrm{6}} ^{\mathrm{9}} {k} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}\right)\:=\frac{\mathrm{13}+\mathrm{17}}{\mathrm{6}}\:=\frac{\mathrm{30}}{\mathrm{6}}=\mathrm{5}\:\Rightarrow \\ $$$${I}\:=\:\mathrm{3}\:+\mathrm{9}\left(\frac{\mathrm{25}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\mathrm{15}\:=\mathrm{18}+\frac{\mathrm{25}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$=\mathrm{18}\:+\mathrm{8}\:\:=\mathrm{26} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 25/Mar/20
Commented by M±th+et£s last updated on 25/Mar/20
$${thank}\:{you}\:{sir}\:{but}\:{can}\:{you}\:{tell}\:{me}\:{where} \\ $$$${i}\:{make}\:{a}\:{mistake} \\ $$
Commented by abdomathmax last updated on 25/Mar/20
$${i}\:{dont}\:{know}\:{why}\:{you}\:{take}\:\mathrm{3},\mathrm{5}\:{and}\:\mathrm{4},\mathrm{5}... \\ $$
Commented by M±th+et£s last updated on 25/Mar/20
$${because}\:{inside}\:{the}\:{brackets}\:{not}\:{the}\:{exponent}\: \\ $$$${is}\:{multiply}\:{by}\:\mathrm{2} \\ $$