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Question Number 122942 by CanovasCamiseros last updated on 21/Nov/20

Commented by CanovasCamiseros last updated on 21/Nov/20

$$\mathit{h}\mathit{e}\mathit{l}\mathit{p}\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}$$

Answered by liberty last updated on 21/Nov/20

$$\underset{0}{\stackrel{1}{\int }}(e^{-x}-e^{-2x})dx={[-e^{-x}+\frac{1}{2}{e}^{-2x}]}_0^1$$$$=(-e^{-1}+\frac{1}{2}{e}^{-2})-(-1+\frac{1}{2})$$$$=\frac{1}{2}-\frac{1}{e}+\frac{1}{2e^2}=\frac{e^2-2e+1}{2e^2}=\frac{1}{2}{\left(\frac{e-1}{e}\right)}^2.$$

Commented by CanovasCamiseros last updated on 21/Nov/20

$$\mathit{T}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}!$$$$$$

Answered by mathmax by abdo last updated on 21/Nov/20

$$\mathrm{A}={\int }_0^1\frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}\mathrm{changement}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}\mathrm{give}$$$$\mathrm{A}={\int }_1^{\mathrm{e}}\frac{\mathrm{t}-1}{{\mathrm{t}}^2}\frac{\mathrm{dt}}{\mathrm{t}}={\int }_1^{\mathrm{e}}\frac{\mathrm{t}-1}{{\mathrm{t}}^3}\mathrm{dt}={\int }_1^{\mathrm{e}}\frac{\mathrm{dt}}{{\mathrm{t}}^2}-{\int }_1^{\mathrm{e}}{\mathrm{t}}^{-3}\mathrm{dt}$$$$={[-\frac{1}{\mathrm{t}}]}_1^{\mathrm{e}}-{\left[\frac{1}{-3+1}{\mathrm{t}}^{-3+1}\right]}_1^{\mathrm{e}}=1-\frac{1}{\mathrm{e}}+\frac{1}{2}{\left[\frac{1}{{\mathrm{t}}^2}\right]}_1^{\mathrm{e}}$$$$=1-\frac{1}{\mathrm{e}}+\frac{1}{2}(1-\frac{1}{{\mathrm{e}}^2})$$

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