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Question Number 62924 by mathmax by abdo last updated on 26/Jun/19
$${find}\:{min}_{\left({a},{b}\right)\in{R}^{\mathrm{2}} } \:\:\:\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \left({ax}+{b}\right)^{\mathrm{2}} {dx} \\ $$
Commented by kaivan.ahmadi last updated on 27/Jun/19
![u=ax+b⇒du=adx (1/a)∫u^2 du=(u^3 /(3a))=(((ax+b)^3 )/(3a))∣_(−1) ^1 =(1/(3a))[(a+b)^3 −(b−a)^3 ]= (1/(3a))[(a^3 +3a^2 b+3ab^2 +b^3 )−(b^3 −3ab^2 +3a^2 b−a^3 )]= (1/(3a))(2a^3 +6ab^2 )=(2/3)a^2 +2b^2](Q62932.png)
$${u}={ax}+{b}\Rightarrow{du}={adx} \\ $$$$\frac{\mathrm{1}}{{a}}\int{u}^{\mathrm{2}} {du}=\frac{{u}^{\mathrm{3}} }{\mathrm{3}{a}}=\frac{\left({ax}+{b}\right)^{\mathrm{3}} }{\mathrm{3}{a}}\mid_{−\mathrm{1}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}{a}}\left[\left({a}+{b}\right)^{\mathrm{3}} −\left({b}−{a}\right)^{\mathrm{3}} \right]= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{a}}\left[\left({a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} \right)−\left({b}^{\mathrm{3}} −\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} {b}−{a}^{\mathrm{3}} \right)\right]= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{a}}\left(\mathrm{2}{a}^{\mathrm{3}} +\mathrm{6}{ab}^{\mathrm{2}} \right)=\frac{\mathrm{2}}{\mathrm{3}}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 27/Jun/19
$${thanks}\:{sir}\:{Ahmadi}. \\ $$
Answered by MJS last updated on 27/Jun/19
![∫^1 _(−1) (ax+b)^2 dx=[(((ax+b)^3 )/(3a))]_(−1) ^1 =(2/3)(a^2 +3b^2 ) the minimum of this is zero with a=b=0](Q62933.png)
$$\underset{−\mathrm{1}} {\int}^{\mathrm{1}} \left({ax}+{b}\right)^{\mathrm{2}} {dx}=\left[\frac{\left({ax}+{b}\right)^{\mathrm{3}} }{\mathrm{3}{a}}\right]_{−\mathrm{1}} ^{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right) \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{this}\:\mathrm{is}\:\mathrm{zero}\:\mathrm{with}\:{a}={b}=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 27/Jun/19
$${thanks}\:{sir}\:{mjs} \\ $$