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

$$Letf\left(x\right)={\int }_2^x\frac{dt}{1+t^6}.$$ $$Provethat:\frac{1}{730}<f\left(3\right)<\frac{1}{65}.$$

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

$$F\left(t\right)=\frac{1}{1+t^6}isadecreasingfunction$$ $$f\left(3\right)={\int }_2^3\frac{dt}{1+t^6}$$ $$f\left(3\right)<F\left(2\right)\times (3-2)=F\left(2\right)=\frac{1}{1+2^6}=\frac{1}{1+64}=\frac{1}{65}$$ $$f\left(3\right)>F\left(3\right)\times (3-2)=F\left(3\right)=\frac{1}{1+3^6}=\frac{1}{1+9\times 9\times 9}=\frac{1}{1+729}=\frac{1}{730}$$ $$\Rightarrow \frac{1}{730}<f\left(3\right)<\frac{1}{65}$$

Commented bymr W last updated on 14/Dec/18

Commented bymr W last updated on 15/Dec/18

$$itisexactasyoucanseeinthediagram.$$ $$thereexistssuchapoint\xi .weknow$$ $$itexists,butwe{don}^{\prime }tknowwhereit$$ $$is.$$ $${\int }_a^{b}F\left(t\right)dt=areaunderthecurvefromatob.$$ $$$$ $$ifF\left(t\right)isdecreasing,{it}^{\prime }sobviousthat$$ $${\int }_a^{b}F\left(t\right)dt<(b-a)F\left(a\right)=rectanglewithmax.height$$ $${\int }_a^{b}F\left(t\right)dt>(b-a)F\left(b\right)=rectanglewithmin.height$$ $$$$ $$ifF\left(t\right)isincreasing,{it}^{\prime }sobviousthat$$ $${\int }_a^{b}F\left(t\right)dt>(b-a)F\left(a\right)$$ $${\int }_a^{b}F\left(t\right)dt<(b-a)F\left(b\right)$$

Commented byrahul 19 last updated on 15/Dec/18

$$Sir,f\left(3\right)=F\left(\xi \right)(3-2)isjustan$$ $$approximation,notexactlyequal,right?$$ $$ThankyouSir!$$

Commented byrahul 19 last updated on 15/Dec/18

Perfect explaination!��

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