Question Number 63372 by necx1 last updated on 03/Jul/19
$${For}\:{what}\:{values}\:{of}\:{a}\:{and}\:{b}\:{will}\:{the} \\ $$$${integral}\:\int_{{a}} ^{{b}} \sqrt{\mathrm{10}−{x}−{x}^{\mathrm{2}} }{dx}\:{be}\:{at} \\ $$$${maximum} \\ $$
Commented by mr W last updated on 03/Jul/19
$${y}=\sqrt{\mathrm{10}−{x}−{x}^{\mathrm{2}} }=\sqrt{\left(\frac{\sqrt{\mathrm{41}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${this}\:{is}\:{a}\:{semicircle}\:{with}\:{radius}\:\frac{\sqrt{\mathrm{41}}}{\mathrm{2}}\:{and} \\ $$$${center}\:\left(−\frac{\mathrm{1}}{\mathrm{2}},\mathrm{0}\right) \\ $$$${max}.\:{value}\:{of}\:{integral}\:{is}\:{the}\:{area} \\ $$$${of}\:{semicircle},\:{i}.{e}. \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{41}}}{\mathrm{2}} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{41}}}{\mathrm{2}} \\ $$$${max}.\:{I}=\frac{\pi}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{41}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{41}\pi}{\mathrm{8}} \\ $$
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...](Q63388.png)
$${let}\:{f}\left({a},{b}\right)\:=\int_{{a}} ^{{b}} \sqrt{\mathrm{10}−{x}−{x}^{\mathrm{2}} }{dx}\:\Rightarrow{f}\left({a},{b}\right)\:=\int_{{a}} ^{{b}} \sqrt{−\left({x}^{\mathrm{2}} +{x}−\mathrm{10}\right)}{dx} \\ $$$$=\int_{{a}} ^{{b}} \sqrt{−\left(\:{x}^{\mathrm{2}} +\mathrm{2}{x}\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{10}\right)}{dx}\:=\int_{{a}} ^{{b}} \sqrt{−\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{10}}{dx} \\ $$$$=\int_{{a}} ^{{b}} \sqrt{\frac{\mathrm{41}}{\mathrm{4}}−\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}\:\:\:{changement}\:\:{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{41}}}{\mathrm{2}}{sint}\:{give} \\ $$$${f}\left({a},{b}\right)\:=\int_{\frac{\mathrm{2}{a}+\mathrm{1}}{\sqrt{\mathrm{41}}}} ^{\frac{\mathrm{2}{b}+\mathrm{1}}{\sqrt{\mathrm{41}}}} \:\:\frac{\sqrt{\mathrm{41}}}{\mathrm{2}}\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}\frac{\sqrt{\mathrm{41}}}{\mathrm{2}}{cost}\:{dt} \\ $$$$=\frac{\mathrm{41}}{\mathrm{4}}\:\int_{\frac{\mathrm{2}{a}+\mathrm{1}}{\sqrt{\mathrm{41}}}} ^{\frac{\mathrm{2}{b}+\mathrm{1}}{\sqrt{\mathrm{41}}}} \:\:{cos}^{\mathrm{2}} {t}\:{dt}\:=\frac{\mathrm{41}}{\mathrm{8}}\:\int_{\frac{\mathrm{2}{a}+\mathrm{1}}{\sqrt{\mathrm{41}}}} ^{\frac{\mathrm{2}{b}+\mathrm{1}}{\sqrt{\mathrm{41}}}} \:\:\:\left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{41}}{\mathrm{8}}\left\{\frac{\mathrm{2}{b}+\mathrm{1}}{\sqrt{\mathrm{41}}}\:−\frac{\mathrm{2}{a}+\mathrm{1}}{\sqrt{\mathrm{41}}}\right\}\:+\frac{\mathrm{41}}{\mathrm{16}}\:\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\frac{\mathrm{2}{a}+\mathrm{1}}{\sqrt{\mathrm{41}}}} ^{\frac{\mathrm{2}{b}+\mathrm{1}}{\sqrt{\mathrm{41}}}} \\ $$$$=\frac{\sqrt{\mathrm{41}}}{\mathrm{4}}\left({b}−{a}\right)\:+\frac{\mathrm{41}}{\mathrm{16}}\left\{\:{sin}\left(\frac{\mathrm{4}{b}+\mathrm{2}}{\sqrt{\mathrm{41}}}\right)−{sin}\left(\frac{\mathrm{4}{a}+\mathrm{2}}{\sqrt{\mathrm{41}}}\right)\right\}\:\:{let}\:{suppose}\:{b}\geqslant{a}\:\:\left({b}\:{fixed}\right) \\ $$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=−\frac{\sqrt{\mathrm{41}}}{\mathrm{4}}\:−\frac{\mathrm{41}}{\mathrm{16}}\:\frac{\mathrm{4}}{\sqrt{\mathrm{41}}}\:{cos}\left(\frac{\mathrm{4}{a}+\mathrm{2}}{\sqrt{\mathrm{41}}}\right)\:=−\frac{\sqrt{\mathrm{41}}}{\mathrm{4}}\:−\frac{\sqrt{\mathrm{41}}}{\mathrm{4}}\:{cos}\left(\frac{\mathrm{4}{a}+\mathrm{2}}{\sqrt{\mathrm{41}}}\right) \\ $$$$=−\frac{\sqrt{\mathrm{41}}}{\mathrm{4}}\left\{\:\mathrm{1}+{cos}\left(\frac{\mathrm{4}{a}+\mathrm{2}}{\sqrt{\mathrm{41}}}\right)\right\}\:=\mathrm{0}\:\Rightarrow{cos}\left(\frac{\mathrm{4}{a}+\mathrm{2}}{\sqrt{\mathrm{41}}}\right)={cos}\left(\pi\right)\:\Rightarrow \\ $$$$\frac{\mathrm{4}{a}+\mathrm{2}}{\sqrt{\mathrm{41}}}\:=\pi\:+\mathrm{2}{k}\pi\:\:{or}\:\:\:\frac{\mathrm{4}{a}+\mathrm{2}}{\sqrt{\mathrm{41}}}\:=−\pi\:+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$$\mathrm{4}{a}+\mathrm{2}\:=\sqrt{\mathrm{41}}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\:{or}\:\:\:\mathrm{4}{a}+\mathrm{2}\:=\sqrt{\mathrm{41}}\left(\mathrm{2}{k}−\mathrm{1}\right)\pi\:\Rightarrow \\ $$$${a}\:=\frac{\sqrt{\mathrm{41}}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi−\mathrm{2}}{\mathrm{4}}\:{or}\:\:\:{a}\:=\frac{\sqrt{\mathrm{41}}\left(\mathrm{2}{k}−\mathrm{1}\right)\pi\:−\mathrm{2}}{\mathrm{4}}\:\:\:\:\:\:\:\:\left({k}\:\in{Z}\right)\:\:{rest}\:{to}\:{dresz}\:{the}\: \\ $$$${variation}\:{of}\:{f}\left({a},{b}\right).....{be}\:{continued}... \\ $$
Answered by MJS last updated on 03/Jul/19
![F(x)=∫(√(10−x−x^2 ))dx ∫_a ^b (√(10−x−x^2 ))dx=F(b)−F(a) the maximum for a is at (d/da)[F(b)−F(a)]=(√(10−a−a^2 ))=0 ⇒ a=−(1/2)±((√(41))/2) the maximum for b is similar at b=−(1/2)±((√(41))/2) b>a ⇒ a=−(1/2)−((√(41))/2)∧b=−(1/2)+((√(41))/2)](Q63374.png)
$${F}\left({x}\right)=\int\sqrt{\mathrm{10}−{x}−{x}^{\mathrm{2}} }{dx} \\ $$$$\underset{{a}} {\overset{{b}} {\int}}\sqrt{\mathrm{10}−{x}−{x}^{\mathrm{2}} }{dx}={F}\left({b}\right)−{F}\left({a}\right) \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{for}\:{a}\:\mathrm{is}\:\mathrm{at} \\ $$$$\frac{{d}}{{da}}\left[{F}\left({b}\right)−{F}\left({a}\right)\right]=\sqrt{\mathrm{10}−{a}−{a}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{a}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{41}}}{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{for}\:{b}\:\mathrm{is}\:\mathrm{similar}\:\mathrm{at} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{41}}}{\mathrm{2}} \\ $$$${b}>{a}\:\Rightarrow\:{a}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{41}}}{\mathrm{2}}\wedge{b}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{41}}}{\mathrm{2}} \\ $$