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Question Number 73730 by TawaTawa last updated on 15/Nov/19

Answered by mind is power last updated on 15/Nov/19

$$f\left(x\right)=e^x-e^c-e^c(x-c)$$$$\Rightarrow f^{\prime }\left(x\right)=e^x-e^c\Rightarrow \{\begin{array}{l}f\left({}^{\prime }x\right)⩾0,\mathrm{\forall }x⩾c\\ f^{\prime }\left(x\right)⩽0,\mathrm{\forall }x⩽c\end{array}$$$$\Rightarrow minf=f\left(c\right)=0\Rightarrow \mathrm{\forall }x\in \mathbb{R},f\left(x\right)⩾0\iff e^x-e^c⩾e^c(x-c)$$$${\int }_0^1{e}^{t^{R+1}}dt$$$$wehaveforc=0\Rightarrow e^x-1⩾1(x-1)\iff e^x⩾x,x=t^{R+2}$$$$\Rightarrow e^{t^{R+2}}⩾t^{R+2}$$$$\Rightarrow {\int }_0^1{e}^{t^{R+2}}⩾{\int }_0^1{t}^{R+2}=\frac{1}{R+3}$$

Commented by TawaTawa last updated on 15/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mind is power last updated on 15/Nov/19

$$y^{\prime }rewelcomsir$$

Answered by mind is power last updated on 15/Nov/19

$${\int }_0^1{e}^{t^{R+2}}dt⩾e^{\frac{1}{R+3}}sorrymypreviousanswerwhenizoomededidnotappear$$$$weusea\left(i\right)forx=t^{R+2},andc=\frac{1}{R+3}$$$$\Rightarrow e^{t^{R+2}}-e^{\frac{1}{R+3}}⩾e^{\frac{1}{R+3}}(t^{R+2}-\frac{1}{R+3})$$$$\Rightarrow {\int }_0^1(e^{t^{R+2}}-e^{\frac{1}{R+3}})dt⩾\int e^{\frac{1}{R+3}}(t^{R+2}-\frac{1}{R+3})dt=e^{\frac{1}{R+3}}{\int }_0^1(t^{R+2}-\frac{1}{R+3})dt=0$$$$\Rightarrow {\int }_0^1(e^{t^{R+2}}-e^{\frac{1}{R+3}})dt⩾0$$$${\int }_0^1(e^{t^{R+2}}-e^{\frac{1}{R+3}})dt={\int }_0^1{e}^{t^{R+2}}dt-e^{\frac{1}{R+3}}⩾0\Rightarrow {\int }_0^1{e}^{t^{R+2}}⩾e^{\frac{1}{R+3}}$$$$$$$$$$

Commented by TawaTawa last updated on 15/Nov/19

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by mind is power last updated on 15/Nov/19

$$y^{\prime }rewelcom$$

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