find min_((a,b)∈R^2 ) ∫_(−1) ^1 (ax+b)^2 dx


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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

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