Question Number 209557 by Spillover last updated on 14/Jul/24
$${Given}\: \\ $$$$\:\int_{\mathrm{2}} ^{\mathrm{4}} \left({ax}^{{n}} +\mathrm{1}\right){dx}=\mathrm{58}\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \left({ax}^{{n}} +\mathrm{1}\right){dx}=\mathrm{10} \\ $$$${find}\:{the}\:{value}\:{of}\:\:{a}\:\:\:{and}\:\:{n} \\ $$
Answered by mr W last updated on 14/Jul/24
![∫_0 ^2 (ax^n +1)dx=[((ax^(n+1) )/(n+1))+x]_0 ^2 =((a2^(n+1) )/(n+1))+2=10 ⇒((a2^(n+1) )/(n+1))=8 ...(i) ∫_0 ^4 (ax^n +1)dx=((a4^(n+1) )/(n+1))+4=10+58=68 ⇒((a4^(n+1) )/(n+1))=64 ...(ii) (ii)/(i): ((4/2))^(n+1) =8 ⇒n=2 ✓ ⇒a=((8×3)/2^3 )=3 ✓](https://www.tinkutara.com/question/Q209558.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}} \left({ax}^{{n}} +\mathrm{1}\right){dx}=\left[\frac{{ax}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+{x}\right]_{\mathrm{0}} ^{\mathrm{2}} =\frac{{a}\mathrm{2}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\mathrm{2}=\mathrm{10} \\ $$$$\Rightarrow\frac{{a}\mathrm{2}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\mathrm{8}\:\:\:…\left({i}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \left({ax}^{{n}} +\mathrm{1}\right){dx}=\frac{{a}\mathrm{4}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\mathrm{4}=\mathrm{10}+\mathrm{58}=\mathrm{68} \\ $$$$\Rightarrow\frac{{a}\mathrm{4}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\mathrm{64}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)/\left({i}\right):\:\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} =\mathrm{8}\:\Rightarrow{n}=\mathrm{2}\:\checkmark \\ $$$$\Rightarrow{a}=\frac{\mathrm{8}×\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }=\mathrm{3}\:\checkmark \\ $$
Commented by Spillover last updated on 14/Jul/24
$${thank}\:{you} \\ $$
Answered by lepuissantcedricjunior last updated on 16/Jul/24
![∫_2 ^4 (ax^n +1)dx=58 ∫_0 ^2 (ax^n +1)dx=10 => { ((∫_2 ^4 (ax^n +1)dx)),((∫_0 ^2 (ax^n +1)dx)) :}=> { ((=[(a/(1+n))x^(n+1) +x]_2 ^4 )),((=[(a/(n+1))x^(n+1) +x]_0 ^2 )) :}=> { ((=((2^(2n+2) a)/(1+n))−((2^(n+1) a)/(n+1))+2)),((=((2^(n+1) a)/(n+1))+2)) :} => { ((((2^(n+1) a(2^(n+1) −1))/(n+1))=56(1))),((((2^(n+1) a)/(n+1))=8(2)remplacons dans (1))) :} =>8(2^(n+1) −1)=56=>2^(n+1) =8=2^3 =>n+1=3=>n=2 deplus ((2^(n+1) a)/(n+1))=8=>(8/3)a=8=>a=3 d′ou′ ⇒{(n;a)}={(2;3)} ............le puissant cedric junior........](https://www.tinkutara.com/question/Q209615.png)
$$\int_{\mathrm{2}} ^{\mathrm{4}} \left(\boldsymbol{{ax}}^{\boldsymbol{{n}}} +\mathrm{1}\right)\boldsymbol{{dx}}=\mathrm{58} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \left(\boldsymbol{{ax}}^{\boldsymbol{{n}}} +\mathrm{1}\right)\boldsymbol{{dx}}=\mathrm{10} \\ $$$$=>\begin{cases}{\int_{\mathrm{2}} ^{\mathrm{4}} \left(\boldsymbol{{ax}}^{\boldsymbol{{n}}} +\mathrm{1}\right)\boldsymbol{{dx}}}\\{\int_{\mathrm{0}} ^{\mathrm{2}} \left(\boldsymbol{{ax}}^{\boldsymbol{{n}}} +\mathrm{1}\right)\boldsymbol{{dx}}}\end{cases}=>\begin{cases}{=\left[\frac{\boldsymbol{{a}}}{\mathrm{1}+\boldsymbol{{n}}}\boldsymbol{{x}}^{\boldsymbol{{n}}+\mathrm{1}} +\boldsymbol{{x}}\right]_{\mathrm{2}} ^{\mathrm{4}} }\\{=\left[\frac{\boldsymbol{{a}}}{\boldsymbol{{n}}+\mathrm{1}}\boldsymbol{{x}}^{\boldsymbol{{n}}+\mathrm{1}} +\boldsymbol{{x}}\right]_{\mathrm{0}} ^{\mathrm{2}} }\end{cases}=>\begin{cases}{=\frac{\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}+\mathrm{2}} \boldsymbol{{a}}}{\mathrm{1}+\boldsymbol{{n}}}−\frac{\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} \boldsymbol{{a}}}{\boldsymbol{{n}}+\mathrm{1}}+\mathrm{2}}\\{=\frac{\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} \boldsymbol{{a}}}{\boldsymbol{{n}}+\mathrm{1}}+\mathrm{2}}\end{cases} \\ $$$$=>\begin{cases}{\frac{\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} \boldsymbol{{a}}\left(\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} −\mathrm{1}\right)}{\boldsymbol{{n}}+\mathrm{1}}=\mathrm{56}\left(\mathrm{1}\right)}\\{\frac{\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} \boldsymbol{{a}}}{\boldsymbol{{n}}+\mathrm{1}}=\mathrm{8}\left(\mathrm{2}\right)\boldsymbol{{remplacons}}\:\boldsymbol{{dans}}\:\left(\mathrm{1}\right)}\end{cases} \\ $$$$=>\mathrm{8}\left(\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} −\mathrm{1}\right)=\mathrm{56}=>\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} =\mathrm{8}=\mathrm{2}^{\mathrm{3}} \\ $$$$=>\boldsymbol{{n}}+\mathrm{1}=\mathrm{3}=>\boldsymbol{{n}}=\mathrm{2} \\ $$$$\boldsymbol{{deplus}}\:\frac{\mathrm{2}^{\boldsymbol{{n}}+\mathrm{1}} \boldsymbol{{a}}}{\boldsymbol{{n}}+\mathrm{1}}=\mathrm{8}=>\frac{\mathrm{8}}{\mathrm{3}}\boldsymbol{{a}}=\mathrm{8}=>\boldsymbol{{a}}=\mathrm{3} \\ $$$$\boldsymbol{{d}}'\boldsymbol{{ou}}'\:\Rightarrow\left\{\left(\boldsymbol{{n}};\boldsymbol{{a}}\right)\right\}=\left\{\left(\mathrm{2};\mathrm{3}\right)\right\} \\ $$$$…………{le}\:{puissant}\:{cedric}\:{junior}…….. \\ $$
Commented by Spillover last updated on 30/Jul/24
$${great} \\ $$