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Question Number 197629 by SANOGO last updated on 24/Sep/23

$${\int }_0^3{e}^{tE\left(t\right)}dt$$

Answered by Mathspace last updated on 24/Sep/23

$$I={\int }_0^3{e}^{t\left[t\right]}dt={\int }_0^1{e}^{t\left[t\right]}+{\int }_1^2{e}^{t\left[t\right]}dt$$$$+{\int }_2^3{e}^{t\left[t\right]}dt$$$$=0+{\int }_1^2{e}^{t}dt+{\int }_2^3{e}^{2t}dt$$$$={\left[e^t\right]}_1^2+{\left[\frac{1}{2}{e}^{2t}\right]}_2^3$$$$=e^2-e+\frac{1}{2}(e^6-e^4)$$

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