For what values of a and b will the integral ∫_a ^b (√(10−x−x^2 ))dx be at maximum


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Question Number 63372 by necx1 last updated on 03/Jul/19

$F o r w h a t v a l u e s o f a a n d b w i l l t h e$ $i n t e g r a l \int_{a}^{b} \sqrt{10 - x - x^{2}} d x b e a t$ $m a x i m u m$
Commented by mr W last updated on 03/Jul/19

$y = \sqrt{10 - x - x^{2}} = \sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x + \frac{1}{2})}^{2}}$ $t h i s i s a s e m i c i r c l e w i t h r a d i u s \frac{\sqrt{41}}{2} a n d$ $c e n t e r (- \frac{1}{2}, 0)$ $m a x . v a l u e o f i n t e g r a l i s t h e a r e a$ $o f s e m i c i r c l e, i . e .$ $a = - \frac{1}{2} - \frac{\sqrt{41}}{2}$ $b = - \frac{1}{2} + \frac{\sqrt{41}}{2}$ $m a x . I = \frac{π}{2} {(\frac{\sqrt{41}}{2})}^{2} = \frac{41 π}{8}$
Commented by mathmax by abdo last updated on 03/Jul/19
$let f(a,b) =∫_a ^b (√(10−x−x^2 ))dx ⇒f(a,b) =∫_a ^b (√(−(x^2 +x−10)))dx =∫_a ^b (√(−( x^2 +2x(1/2)+(1/4)−(1/4)−10)))dx =∫_a ^b (√(−(x+(1/2))^2 +(1/4)+10))dx =∫_a ^b (√(((41)/4)−(x+(1/2))^2 ))dx changement x+(1/2)=((√(41))/2)sint give f(a,b) =∫_((2a+1)/(√(41))) ^((2b+1)/(√(41))) ((√(41))/2) (√(1−sin^2 t))((√(41))/2)cost dt =((41)/4) ∫_((2a+1)/(√(41))) ^((2b+1)/(√(41))) cos^2 t dt =((41)/8) ∫_((2a+1)/(√(41))) ^((2b+1)/(√(41))) (1+cos(2t))dt =((41)/8){((2b+1)/(√(41))) −((2a+1)/(√(41)))} +((41)/(16)) [sin(2t)]_((2a+1)/(√(41))) ^((2b+1)/(√(41))) =((√(41))/4)(b−a) +((41)/(16)){ sin(((4b+2)/(√(41))))−sin(((4a+2)/(√(41))))} let suppose b≥a (b fixed) (∂f/∂a)(a,b) =−((√(41))/4) −((41)/(16)) (4/(√(41))) cos(((4a+2)/(√(41)))) =−((√(41))/4) −((√(41))/4) cos(((4a+2)/(√(41)))) =−((√(41))/4){ 1+cos(((4a+2)/(√(41))))} =0 ⇒cos(((4a+2)/(√(41))))=cos(π) ⇒ ((4a+2)/(√(41))) =π +2kπ or ((4a+2)/(√(41))) =−π +2kπ ⇒ 4a+2 =(√(41))(2k+1)π or 4a+2 =(√(41))(2k−1)π ⇒ a =(((√(41))(2k+1)π−2)/4) or a =(((√(41))(2k−1)π −2)/4) (k ∈Z) rest to dresz the variation of f(a,b).....be continued...$
$l e t f (a, b) = \int_{a}^{b} \sqrt{10 - x - x^{2}} d x \Rightarrow f (a, b) = \int_{a}^{b} \sqrt{- (x^{2} + x - 10)} d x$ $= \int_{a}^{b} \sqrt{- (x^{2} + 2 x \frac{1}{2} + \frac{1}{4} - \frac{1}{4} - 10)} d x = \int_{a}^{b} \sqrt{- {(x + \frac{1}{2})}^{2} + \frac{1}{4} + 10} d x$ $= \int_{a}^{b} \sqrt{\frac{41}{4} - {(x + \frac{1}{2})}^{2}} d x c h a n g e m e n t x + \frac{1}{2} = \frac{\sqrt{41}}{2} s i n t g i v e$ $f (a, b) = \int_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}} \frac{\sqrt{41}}{2} \sqrt{1 - {s i n}^{2} t} \frac{\sqrt{41}}{2} c o s t d t$ $= \frac{41}{4} \int_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}} {c o s}^{2} t d t = \frac{41}{8} \int_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}} (1 + c o s (2 t)) d t$ $= \frac{41}{8} {\frac{2 b + 1}{\sqrt{41}} - \frac{2 a + 1}{\sqrt{41}}} + \frac{41}{16} {[s i n (2 t)]}_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}}$ $= \frac{\sqrt{41}}{4} (b - a) + \frac{41}{16} {s i n (\frac{4 b + 2}{\sqrt{41}}) - s i n (\frac{4 a + 2}{\sqrt{41}})} l e t s u p p o s e b ⩾ a (b f i x e d)$ $\frac{\partial f}{\partial a} (a, b) = - \frac{\sqrt{41}}{4} - \frac{41}{16} \frac{4}{\sqrt{41}} c o s (\frac{4 a + 2}{\sqrt{41}}) = - \frac{\sqrt{41}}{4} - \frac{\sqrt{41}}{4} c o s (\frac{4 a + 2}{\sqrt{41}})$ $= - \frac{\sqrt{41}}{4} {1 + c o s (\frac{4 a + 2}{\sqrt{41}})} = 0 \Rightarrow c o s (\frac{4 a + 2}{\sqrt{41}}) = c o s (π) \Rightarrow$ $\frac{4 a + 2}{\sqrt{41}} = π + 2 k π o r \frac{4 a + 2}{\sqrt{41}} = - π + 2 k π \Rightarrow$ $4 a + 2 = \sqrt{41} (2 k + 1) π o r 4 a + 2 = \sqrt{41} (2 k - 1) π \Rightarrow$ $a = \frac{\sqrt{41} (2 k + 1) π - 2}{4} o r a = \frac{\sqrt{41} (2 k - 1) π - 2}{4} (k \in Z) r e s t t o d r e s z t h e$ $v a r i a t i o n o f f (a, b) . . . . . b e c o n t i n u e d . . .$
	Answered by MJS last updated on 03/Jul/19

	$F (x) = \int \sqrt{10 - x - x^{2}} d x$ $\underset{a}{\int^{b}} \sqrt{10 - x - x^{2}} d x = F (b) - F (a)$ $the maximum for a is at$ $\frac{d}{d a} [F (b) - F (a)] = \sqrt{10 - a - a^{2}} = 0 \Rightarrow a = - \frac{1}{2} \pm \frac{\sqrt{41}}{2}$ $the maximum for b is similar at$ $b = - \frac{1}{2} \pm \frac{\sqrt{41}}{2}$ $b > a \Rightarrow a = - \frac{1}{2} - \frac{\sqrt{41}}{2} \land b = - \frac{1}{2} + \frac{\sqrt{41}}{2}$
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