Question Number 97117 by bobhans last updated on 06/Jun/20
![lim_(h→0) [ (1/h) ∫_2 ^(2+2h) (√(t^2 +2)) dt ]](Q97117.png)
$$\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{h}}\:\underset{\mathrm{2}} {\overset{\mathrm{2}+\mathrm{2h}} {\int}}\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{2}}\:\mathrm{dt}\:\right]\: \\ $$
Answered by abdomathmax last updated on 06/Jun/20
![let use hospital theorem ⇒ lim_(h→0) ((∫_2 ^(2+2h) (√(t^2 +2))dt)/h) =lim_(h→0) 2(√((2+2h)^2 +2))=2(√(4+2))=2(√6) another way ∃c ∈]2,2+2h[ /A(h)= (1/h)∫_2 ^(2+2h) (√(t^2 +2))dt =(1/h)(√(c^2 +2))∫_2 ^(2+2h) dt =2(√(c^2 +2)) (h→0) ⇒c→2 ⇒ lim_(h→0) A(h) =2(√6)](Q97123.png)
$$\mathrm{let}\:\mathrm{use}\:\mathrm{hospital}\:\mathrm{theorem}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\frac{\int_{\mathrm{2}} ^{\mathrm{2}+\mathrm{2h}} \sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{2}}\mathrm{dt}}{\mathrm{h}} \\ $$$$=\mathrm{lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\:\:\mathrm{2}\sqrt{\left(\mathrm{2}+\mathrm{2h}\right)^{\mathrm{2}} \:+\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{4}+\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\left.\mathrm{another}\:\mathrm{way}\:\exists\mathrm{c}\:\in\right]\mathrm{2},\mathrm{2}+\mathrm{2h}\left[\:/\mathrm{A}\left(\mathrm{h}\right)=\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{h}}\int_{\mathrm{2}} ^{\mathrm{2}+\mathrm{2h}} \sqrt{\mathrm{t}^{\mathrm{2}} \:+\mathrm{2}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{h}}\sqrt{\mathrm{c}^{\mathrm{2}} \:+\mathrm{2}}\int_{\mathrm{2}} ^{\mathrm{2}+\mathrm{2h}} \:\mathrm{dt} \\ $$$$=\mathrm{2}\sqrt{\mathrm{c}^{\mathrm{2}} \:+\mathrm{2}}\:\:\:\left(\mathrm{h}\rightarrow\mathrm{0}\right)\:\Rightarrow\mathrm{c}\rightarrow\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{h}\rightarrow\mathrm{0}} \:\mathrm{A}\left(\mathrm{h}\right)\:=\mathrm{2}\sqrt{\mathrm{6}} \\ $$
Commented by bobhans last updated on 07/Jun/20
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mathmax by abdo last updated on 07/Jun/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$