solve-0-1-1-x-6-dx-mr-abbo-your-question-

Question Number 111039 by mathdave last updated on 01/Sep/20

$${solve}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{6}} }{dx} \\ $$$${mr}\:\:{abbo}\:{your}\:{question}\: \\ $$

Answered by mathdave last updated on 01/Sep/20

let I=∫_0 ^1 (√(1+x^6 ))dx let u=x^6 I=(1/6)∫_0 ^1 u^(−(5/6)) (1+u)^(1/2) du by hypergeometric generalization formula β(b,c−b)2f_1 (a,b,c:z)=∫_0 ^1 x^(b−1) (1−x)^(c−b−1) (1−zx)^(−a) I=(1/6)∫_0 ^1 u^((1/6)−1) (1−u)^((7/6)−(1/6)−1) (1−(−u))^(−(−(1/2))) I=((β((1/6)′1))/6)2f_1 (−(1/2),(1/6),(7/6),:−1)=1.06409 ∵∫_0 ^1 (√(1+x^6 ))dx=1.06409 mathdave(01/09/2020)

$${let}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{6}} }{dx} \\ $$$${let}\:{u}={x}^{\mathrm{6}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{5}}{\mathrm{6}}} \left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {du} \\ $$$${by}\:{hypergeometric}\:{generalization}\:{formula} \\ $$$$\beta\left({b},{c}−{b}\right)\mathrm{2}{f}_{\mathrm{1}} \left({a},{b},{c}:{z}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{b}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{c}−{b}−\mathrm{1}} \left(\mathrm{1}−{zx}\right)^{−{a}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{7}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \left(\mathrm{1}−\left(−{u}\right)\right)^{−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${I}=\frac{\beta\left(\frac{\mathrm{1}}{\mathrm{6}}'\mathrm{1}\right)}{\mathrm{6}}\mathrm{2}{f}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{7}}{\mathrm{6}},:−\mathrm{1}\right)=\mathrm{1}.\mathrm{06409} \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{6}} }{dx}=\mathrm{1}.\mathrm{06409} \\ $$$${mathdave}\left(\mathrm{01}/\mathrm{09}/\mathrm{2020}\right) \\ $$

Commented by abdomsup last updated on 01/Sep/20

$${sorry}\:{i}\:{hsve}\:{red}\:{it}\:{by}\:{eroor}\:{thank} \\ $$$${you}\:{for}\:{the}\:{answer} \\ $$

Commented by Rasheed.Sindhi last updated on 02/Sep/20

$${Only}\:{to}\:{remove}\:{red}\:{mark}\:{I}\:{clicked} \\ $$$${like}\:{button}.{Actually}\:{I}\:{haven}'{t} \\ $$$${read}\:{the}\:{post}. \\ $$

Commented by mathdave last updated on 02/Sep/20

$${i}\:{dont}\:{really}\:{get}\:{what}\:{you}\:{people}\:{are} \\ $$$${saying} \\ $$

Commented by Rasheed.Sindhi last updated on 02/Sep/20

$${mathdave}\:{sir} \\ $$$${I}\:{wanted}\:{to}\:{say}\:{only}\:{that}, \\ $$$${mathmax}\:{sir}\:{unintetionally}\: \\ $$$${marked}\:{your}\:{answer}\:{red}\:{and}\:{I} \\ $$$${wanted}\:{to}\:{remove}\:{that}\:{red}\:{mark}. \\ $$$${so}\:{I}\:{without}\:{reading}\:{your} \\ $$$${answer}\:{clicked}\:{on}\:{like}-{button}\left(\heartsuit\right) \\ $$$${for}\:{removing}\:{red}\:{mark}. \\ $$$$\:\:{In}\:{this}\:{app}\:{if}\:{a}\:{person}\:{marks} \\ $$$${red}\:{a}\:{post}\:{and}\:{another}\:{person} \\ $$$$'{likes}'\:{that}\:{post}\:{then}\:{red}\:{mark} \\ $$$${clears}. \\ $$

Commented by mathdave last updated on 02/Sep/20