Question Number 122942 by CanovasCamiseros last updated on 21/Nov/20 | ||
Commented by CanovasCamiseros last updated on 21/Nov/20 | ||
$$\boldsymbol{{help}}\:\boldsymbol{{please}} \\ $$ | ||
Answered by liberty last updated on 21/Nov/20 | ||
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left({e}^{−{x}} −{e}^{−\mathrm{2}{x}} \right)\:{dx}\:=\:\left[−{e}^{−{x}} +\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} \:\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:=\:\left(−{e}^{−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}} \right)−\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}{e}^{\mathrm{2}} }\:=\:\frac{{e}^{\mathrm{2}} −\mathrm{2}{e}+\mathrm{1}}{\mathrm{2}{e}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{e}−\mathrm{1}}{{e}}\right)^{\mathrm{2}} . \\ $$ | ||
Commented by CanovasCamiseros last updated on 21/Nov/20 | ||
$$\boldsymbol{{Thanks}}\:\boldsymbol{{sir}}! \\ $$$$ \\ $$ | ||
Answered by mathmax by abdo last updated on 21/Nov/20 | ||
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2x}} }\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{e}^{\mathrm{x}} \:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{A}\:=\int_{\mathrm{1}} ^{\mathrm{e}} \:\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\frac{\mathrm{dt}}{\mathrm{t}}\:=\int_{\mathrm{1}} ^{\mathrm{e}} \:\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}^{\mathrm{3}} }\mathrm{dt}\:=\int_{\mathrm{1}} ^{\mathrm{e}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }−\int_{\mathrm{1}} ^{\mathrm{e}} \:\mathrm{t}^{−\mathrm{3}} \:\mathrm{dt} \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{t}}\right]_{\mathrm{1}} ^{\mathrm{e}} \:−\left[\frac{\mathrm{1}}{−\mathrm{3}+\mathrm{1}}\mathrm{t}^{−\mathrm{3}+\mathrm{1}} \right]_{\mathrm{1}} ^{\mathrm{e}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right]_{\mathrm{1}} ^{\mathrm{e}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2}} }\right) \\ $$ | ||