Question Number 197629 by SANOGO last updated on 24/Sep/23
$$\int_{\mathrm{0}} ^{\mathrm{3}} {e}^{{tE}\left({t}\right)} \:{dt} \\ $$
Answered by Mathspace last updated on 24/Sep/23
![I=∫_0 ^3 e^(t[t]) dt=∫_0 ^1 e^(t[t]) +∫_1 ^2 e^(t[t]) dt +∫_2 ^3 e^(t[t]) dt =0 +∫_1 ^2 e^t dt+∫_2 ^3 e^(2t) dt =[e^t ]_1 ^2 +[(1/2)e^(2t) ]_2 ^3 =e^2 −e +(1/2)(e^6 −e^4 )](https://www.tinkutara.com/question/Q197633.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{3}} \:{e}^{{t}\left[{t}\right]} {dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{t}\left[{t}\right]} +\int_{\mathrm{1}} ^{\mathrm{2}} {e}^{{t}\left[{t}\right]} {dt} \\ $$$$+\int_{\mathrm{2}} ^{\mathrm{3}} {e}^{{t}\left[{t}\right]} {dt} \\ $$$$=\mathrm{0}\:+\int_{\mathrm{1}} ^{\mathrm{2}} {e}^{{t}} {dt}+\int_{\mathrm{2}} ^{\mathrm{3}} {e}^{\mathrm{2}{t}} {dt} \\ $$$$=\left[{e}^{{t}} \right]_{\mathrm{1}} ^{\mathrm{2}} +\left[\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{t}} \right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$={e}^{\mathrm{2}} −{e}\:+\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{6}} −{e}^{\mathrm{4}} \right) \\ $$