Question Number 221902 by alvan545 last updated on 12/Jun/25
Answered by mahdipoor last updated on 12/Jun/25
![x−5=u dx=du x=10≡u=5 x=0≡u−5 ∫_(−5) ^( 5) (u+u^2 +u^3 )du=(u^2 /2)+(u^3 /3)+(u^4 /4)]_(−5) ^5](https://www.tinkutara.com/question/Q221904.png)
$${x}−\mathrm{5}={u}\:\:\:\:\:\:\:{dx}={du} \\ $$$${x}=\mathrm{10}\equiv{u}=\mathrm{5}\:\:\:\:\:\:{x}=\mathrm{0}\equiv{u}−\mathrm{5} \\ $$$$\left.\int_{−\mathrm{5}} ^{\:\mathrm{5}} \left({u}+{u}^{\mathrm{2}} +{u}^{\mathrm{3}} \right){du}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}+\frac{{u}^{\mathrm{4}} }{\mathrm{4}}\right]_{−\mathrm{5}} ^{\mathrm{5}} \\ $$
Answered by Frix last updated on 12/Jun/25
![Sophisticated answer: ∫_0 ^(2a) (x−a)^(2n−1) dx =^([t=x−a]) =∫_(−a) ^a t^(2n−1) dt=(1/(2n))[t^(2n) ]_(−a) ^a =0 ∫_0 ^(2a) (x−a)^(2n) dx =^([t=x−a]) =∫_(−a) ^a t^(2n) dt=(1/(2n+1))[t^(2n+1) ]_(−a) ^a =((2a^(2n+1) )/(2n+1)) ⇒ ∫_0 ^(10) ((x−5)^1 +(x−5)^2 +(x−5)^3 )dx= =∫_0 ^(10) (x−5)^2 dx=∫_(−5) ^5 t^2 dt=(1/3)[t^3 ]_(−5) ^5 =((250)/3)](https://www.tinkutara.com/question/Q221905.png)
$$\mathrm{Sophisticated}\:\mathrm{answer}: \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}{a}} {\int}}\left({x}−{a}\right)^{\mathrm{2}{n}−\mathrm{1}} {dx}\:\overset{\left[{t}={x}−{a}\right]} {=} \\ $$$$\:\:\:\:\:=\underset{−{a}} {\overset{{a}} {\int}}{t}^{\mathrm{2}{n}−\mathrm{1}} {dt}=\frac{\mathrm{1}}{\mathrm{2}{n}}\left[{t}^{\mathrm{2}{n}} \right]_{−{a}} ^{{a}} =\mathrm{0} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}{a}} {\int}}\left({x}−{a}\right)^{\mathrm{2}{n}} {dx}\:\overset{\left[{t}={x}−{a}\right]} {=} \\ $$$$\:\:\:\:\:=\underset{−{a}} {\overset{{a}} {\int}}{t}^{\mathrm{2}{n}} {dt}=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left[{t}^{\mathrm{2}{n}+\mathrm{1}} \right]_{−{a}} ^{{a}} =\frac{\mathrm{2}{a}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{10}} {\int}}\left(\left({x}−\mathrm{5}\right)^{\mathrm{1}} +\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({x}−\mathrm{5}\right)^{\mathrm{3}} \right){dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{10}} {\int}}\left({x}−\mathrm{5}\right)^{\mathrm{2}} {dx}=\underset{−\mathrm{5}} {\overset{\mathrm{5}} {\int}}{t}^{\mathrm{2}} {dt}=\frac{\mathrm{1}}{\mathrm{3}}\left[{t}^{\mathrm{3}} \right]_{−\mathrm{5}} ^{\mathrm{5}} =\frac{\mathrm{250}}{\mathrm{3}} \\ $$
Commented by alvan545 last updated on 12/Jun/25
$$ \\ $$100/100, thankyouu
Commented by timmike last updated on 12/Jun/25
I didn't understand ur solution pls ����