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Question Number 50186 by rahul 19 last updated on 14/Dec/18

Let f(x)= ∫_2 ^( x) (dt/(1+t^6 )). Prove that : (1/(730))<f(3)<(1/(65)).

$${Let}\:{f}\left({x}\right)=\:\int_{\mathrm{2}} ^{\:{x}} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{6}} }. \\ $$ $${Prove}\:{that}\:\::\:\frac{\mathrm{1}}{\mathrm{730}}<{f}\left(\mathrm{3}\right)<\frac{\mathrm{1}}{\mathrm{65}}. \\ $$

Answered by mr W last updated on 16/Dec/18

F(t)=(1/(1+t^6 )) is a decreasing function f(3)=∫_2 ^3 (dt/(1+t^6 )) f(3)<F(2)×(3−2)=F(2)=(1/(1+2^6 ))=(1/(1+64))=(1/(65)) f(3)>F(3)×(3−2)=F(3)=(1/(1+3^6 ))=(1/(1+9×9×9))=(1/(1+729))=(1/(730)) ⇒(1/(730))<f(3)<(1/(65))

$${F}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{6}} }\:{is}\:{a}\:{decreasing}\:{function} \\ $$ $${f}\left(\mathrm{3}\right)=\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{6}} } \\ $$ $${f}\left(\mathrm{3}\right)<{F}\left(\mathrm{2}\right)×\left(\mathrm{3}−\mathrm{2}\right)={F}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}^{\mathrm{6}} }=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{64}}=\frac{\mathrm{1}}{\mathrm{65}} \\ $$ $${f}\left(\mathrm{3}\right)>{F}\left(\mathrm{3}\right)×\left(\mathrm{3}−\mathrm{2}\right)={F}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}^{\mathrm{6}} }=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{9}×\mathrm{9}×\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{729}}=\frac{\mathrm{1}}{\mathrm{730}} \\ $$ $$\Rightarrow\frac{\mathrm{1}}{\mathrm{730}}<{f}\left(\mathrm{3}\right)<\frac{\mathrm{1}}{\mathrm{65}} \\ $$

Commented bymr W last updated on 14/Dec/18

Commented bymr W last updated on 15/Dec/18

it is exact as you can see in the diagram. there exists such a point ξ. we know it exists, but we don′t know where it is. ∫_a ^b F(t)dt=area under the curve from a to b. if F(t) is decreasing, it′s obvious that ∫_a ^b F(t)dt<(b−a)F(a)=rectangle with max. height ∫_a ^b F(t)dt>(b−a)F(b)=rectangle with min. height if F(t) is increasing, it′s obvious that ∫_a ^b F(t)dt>(b−a)F(a) ∫_a ^b F(t)dt<(b−a)F(b)

$${it}\:{is}\:{exact}\:{as}\:{you}\:{can}\:{see}\:{in}\:{the}\:{diagram}. \\ $$ $${there}\:{exists}\:{such}\:{a}\:{point}\:\xi.\:\:{we}\:{know} \\ $$ $${it}\:{exists},\:{but}\:{we}\:{don}'{t}\:{know}\:{where}\:{it} \\ $$ $${is}. \\ $$ $$\int_{{a}} ^{{b}} {F}\left({t}\right){dt}={area}\:{under}\:{the}\:{curve}\:{from}\:{a}\:{to}\:{b}. \\ $$ $$ \\ $$ $${if}\:{F}\left({t}\right)\:{is}\:{decreasing},\:{it}'{s}\:{obvious}\:{that} \\ $$ $$\int_{{a}} ^{{b}} {F}\left({t}\right){dt}<\left({b}−{a}\right){F}\left({a}\right)={rectangle}\:{with}\:{max}.\:{height} \\ $$ $$\int_{{a}} ^{{b}} {F}\left({t}\right){dt}>\left({b}−{a}\right){F}\left({b}\right)={rectangle}\:{with}\:{min}.\:{height} \\ $$ $$ \\ $$ $${if}\:{F}\left({t}\right)\:{is}\:{increasing},\:{it}'{s}\:{obvious}\:{that} \\ $$ $$\int_{{a}} ^{{b}} {F}\left({t}\right){dt}>\left({b}−{a}\right){F}\left({a}\right) \\ $$ $$\int_{{a}} ^{{b}} {F}\left({t}\right){dt}<\left({b}−{a}\right){F}\left({b}\right) \\ $$

Commented byrahul 19 last updated on 15/Dec/18

Sir, f(3)=F(ξ)(3−2) is just an approximation,not exactly equal,right? Thank you Sir!

$${Sir},\:{f}\left(\mathrm{3}\right)={F}\left(\xi\right)\left(\mathrm{3}−\mathrm{2}\right)\:{is}\:{just}\:{an}\: \\ $$ $${approximation},{not}\:{exactly}\:{equal},{right}? \\ $$ $${Thank}\:{you}\:{Sir}! \\ $$

Commented byrahul 19 last updated on 15/Dec/18

Perfect explaination!��

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