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Question Number 199385 by hardmath last updated on 02/Nov/23

Find: Ω = ∫_0 ^( 1) x^(15) (√(1 + 3x^8 )) dx = ?

$$\mathrm{Find}: \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\mathrm{15}} \:\sqrt{\mathrm{1}\:+\:\mathrm{3x}^{\mathrm{8}} }\:\mathrm{dx}\:=\:? \\ $$

Answered by witcher3 last updated on 02/Nov/23

x^(15) =x^(7.) x^8 =(1/8)x^8 d(x^8 ) ∫u(√(1+u))=∫(1+u)^(3/2) −(1+u)^(1/2) du....

$$\mathrm{x}^{\mathrm{15}} =\mathrm{x}^{\mathrm{7}.} \mathrm{x}^{\mathrm{8}} =\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{8}} \mathrm{d}\left(\mathrm{x}^{\mathrm{8}} \right) \\ $$$$\int\mathrm{u}\sqrt{\mathrm{1}+\mathrm{u}}=\int\left(\mathrm{1}+\mathrm{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{1}+\mathrm{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{du}.... \\ $$

Commented by hardmath last updated on 02/Nov/23

thank you ser, but how

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{ser},\:\mathrm{but}\:\mathrm{how} \\ $$

Commented by witcher3 last updated on 02/Nov/23

u=x^8 ⇔(1/8)∫_0 ^1 u(√(1+3u)).du =(1/(24))∫_0 ^1 (3u+1−1)(1+3u)^(1/2) du =(1/(24)){∫_0 ^1 (3u+1)^(3/2) du−∫_0 ^1 (1+3u)^(1/2) du =(1/(24))[(1/(3((3/2)+1)))(3u+1)^(5/2) −(1/(3((3/2))))(1+3u)^(3/2) ]_0 ^1 =(1/(24))[(2/(15))(2)^5 −(2/(15))−(2/9)2^3 +(2/9)]

$$\mathrm{u}=\mathrm{x}^{\mathrm{8}} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}\sqrt{\mathrm{1}+\mathrm{3u}}.\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3u}+\mathrm{1}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{3u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3u}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{du}−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{3u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{du}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}\left[\frac{\mathrm{1}}{\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)}\left(\mathrm{3u}+\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\left(\mathrm{1}+\mathrm{3u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}\left[\frac{\mathrm{2}}{\mathrm{15}}\left(\mathrm{2}\right)^{\mathrm{5}} −\frac{\mathrm{2}}{\mathrm{15}}−\frac{\mathrm{2}}{\mathrm{9}}\mathrm{2}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{9}}\right] \\ $$

Commented by York12 last updated on 03/Nov/23

Where are you from

$$\mathrm{Where}\:\mathrm{are}\:\mathrm{you}\:\mathrm{from}\: \\ $$

Commented by witcher3 last updated on 03/Nov/23

i m from france

$$\mathrm{i}\:\mathrm{m}\:\mathrm{from}\:\mathrm{france} \\ $$

Commented by York12 last updated on 03/Nov/23

ur speaking arabic lol

$$\mathrm{ur}\:\mathrm{speaking}\:\mathrm{arabic}\:\mathrm{lol} \\ $$

Commented by witcher3 last updated on 03/Nov/23

yes i speak arabic im algerian livin in France

$$\mathrm{yes}\:\mathrm{i}\:\mathrm{speak}\:\mathrm{arabic}\:\mathrm{im}\:\mathrm{algerian}\:\mathrm{livin}\:\mathrm{in}\:\mathrm{France} \\ $$

Commented by York12 last updated on 03/Nov/23

hmmmm okay nice to meet you :) , I am muslim too.

$$\left.\mathrm{hmmmm}\:\mathrm{okay}\:\mathrm{nice}\:\mathrm{to}\:\mathrm{meet}\:\mathrm{you}\::\right)\:,\:\mathrm{I}\:\mathrm{am}\:\mathrm{muslim} \\ $$$$\mathrm{too}. \\ $$

Commented by York12 last updated on 03/Nov/23

wow ur algerian , nice to hear that , I am saudi

$$\mathrm{wow}\:\mathrm{ur}\:\mathrm{algerian}\:,\:\mathrm{nice}\:\mathrm{to}\:\mathrm{hear}\:\mathrm{that}\:,\:\mathrm{I}\:\mathrm{am}\:\mathrm{saudi} \\ $$

Commented by Frix last updated on 03/Nov/23

I once had a student who stated “I learn Pythagoras just for school because I have to. All pagans and all their friends go to hell.”

$$\mathrm{I}\:\mathrm{once}\:\mathrm{had}\:\mathrm{a}\:\mathrm{student}\:\mathrm{who}\:\mathrm{stated}\:``\mathrm{I}\:\mathrm{learn} \\ $$$$\mathrm{Pythagoras}\:\mathrm{just}\:\mathrm{for}\:\mathrm{school}\:\mathrm{because}\:\mathrm{I}\:\mathrm{have} \\ $$$$\mathrm{to}.\:\mathrm{All}\:\mathrm{pagans}\:\mathrm{and}\:\mathrm{all}\:\mathrm{their}\:\mathrm{friends}\:\mathrm{go}\:\mathrm{to} \\ $$$$\mathrm{hell}.'' \\ $$

Commented by York12 last updated on 03/Nov/23

LMAO

$$\mathrm{LMAO} \\ $$

Commented by witcher3 last updated on 04/Nov/23

nice To meet You yorke selam alaykoum

$$\mathrm{nice}\:\mathrm{To}\:\mathrm{meet}\:\mathrm{You}\:\mathrm{yorke} \\ $$$$\mathrm{selam}\:\mathrm{alaykoum} \\ $$

Commented by witcher3 last updated on 04/Nov/23

lol frix

$$\mathrm{lol}\:\mathrm{frix} \\ $$

Commented by York12 last updated on 04/Nov/23

alaykom alsalam nice to meet you too my name is YUSUF btw

$$\mathrm{alaykom}\:\mathrm{alsalam}\:\mathrm{nice}\:\mathrm{to}\:\mathrm{meet}\:\mathrm{you}\:\mathrm{too} \\ $$$$\mathrm{my}\:\mathrm{name}\:\mathrm{is}\:\mathrm{YUSUF}\:\mathrm{btw} \\ $$

Answered by Frix last updated on 03/Nov/23

∫x^(15) (√(3x^8 +1))dx =^(t=3x^8 +1) =(1/(72))∫(t^(3/2) −t^(1/2) )dt= =(t^(5/2) /(180))−(t^(3/2) /(108))=((t^(3/2) (3t−5))/(540))= =(((3x^8 +1)^(3/2) (9x^8 −2))/(540)) Ω=((29)/(270))

$$\int{x}^{\mathrm{15}} \sqrt{\mathrm{3}{x}^{\mathrm{8}} +\mathrm{1}}{dx}\:\overset{{t}=\mathrm{3}{x}^{\mathrm{8}} +\mathrm{1}} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{72}}\int\left({t}^{\frac{\mathrm{3}}{\mathrm{2}}} −{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \right){dt}= \\ $$$$=\frac{{t}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{180}}−\frac{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{108}}=\frac{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{3}{t}−\mathrm{5}\right)}{\mathrm{540}}= \\ $$$$=\frac{\left(\mathrm{3}{x}^{\mathrm{8}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{9}{x}^{\mathrm{8}} −\mathrm{2}\right)}{\mathrm{540}} \\ $$$$\Omega=\frac{\mathrm{29}}{\mathrm{270}} \\ $$

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