find-min-a-b-R-2-1-1-ax-b-2-dx-

Question Number 62924 by mathmax by abdo last updated on 26/Jun/19

f i n d {m i n}_{(a, b) \in R^{2}} \int_{- 1}^{1} {(a x + b)}^{2} d x

Commented by kaivan.ahmadi last updated on 27/Jun/19

u = a x + b \Rightarrow d u = a d x

\frac{1}{a} \int u^{2} d u = \frac{u^{3}}{3 a} = \frac{{(a x + b)}^{3}}{3 a} ∣_{- 1}^{1} = \frac{1}{3 a} [{(a + b)}^{3} - {(b - a)}^{3}] =

\frac{1}{3 a} [(a^{3} + 3 a^{2} b + 3 {a b}^{2} + b^{3}) - (b^{3} - 3 {a b}^{2} + 3 a^{2} b - a^{3})] =

\frac{1}{3 a} (2 a^{3} + 6 {a b}^{2}) = \frac{2}{3} a^{2} + 2 b^{2}

Commented by mathmax by abdo last updated on 27/Jun/19

t h a n k s s i r A h m a d i .

Answered by MJS last updated on 27/Jun/19

{\int_{- 1}}^{1} {(a x + b)}^{2} d x = {[\frac{{(a x + b)}^{3}}{3 a}]}_{- 1}^{1} = \frac{2}{3} (a^{2} + 3 b^{2})

the minimum of this is zero with a = b = 0

Commented by mathmax by abdo last updated on 27/Jun/19

t h a n k s s i r m j s