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Question Number 31074 by abdo imad last updated on 02/Mar/18
find ∫_a ^b (√((b−x)(x−a))) dx with a<b .then find ∫_1 ^(√2) (√(((√2) −x)(x−1))) dx.
$${find}\:\:\int_{{a}} ^{{b}} \:\sqrt{\left({b}−{x}\right)\left({x}−{a}\right)}\:{dx}\:{with}\:{a}<{b}\:.{then}\:{find}\: \\ $$$$\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \sqrt{\left(\sqrt{\mathrm{2}}\:−{x}\right)\left({x}−\mathrm{1}\right)}\:{dx}. \\ $$
Commented by abdo imad last updated on 09/Mar/18
let put I= ∫_a ^b (√((b−x)(x−a))) dx the ch.x=((a−b)/2)t + ((a+b)/2) b−x=b−((a+b)/2) −((a−b)/2)t=((b−a)/2) −((a−b)/2)t=((b−a)/2)(1+t) x−a=((a−b)/2)t +((a+b)/2) −a= ((a−b)/2)t +((b−a)/2)=((b−a)/2)(1−t) and we have ((a−b)/2)t=x−((a+b)/2) ⇒t=((2(x−((a+b)/2)))/(a−b))=((2x −a−b)/(a−b)) x=a ⇒t=1 and x=b ⇒t=−1 ⇒ I = −∫_(−1) ^(1 ) (√(((b−a)/2)(1+t)((b−a)/2)(1−t))) ((a−b)/2)dt I=(((b−a)^2 )/4) ∫_(−1) ^1 (√(1−t^2 )) dt= (((b−a)^2 )/2) ∫_0 ^1 (√(1−t^2 )) dt ch.t=sinθ give ∫_0 ^1 (√(1−t^2 )) dt=∫_0 ^(π/2) cosθ cosθdθ=∫_0 ^(π/2) cos^2 θ dθ =(1/2) ∫_0 ^(π/2) (1+cos(2θ)dθ = (π/4) ⇒ I= (((b−a)^2 )/8) .then let take a=1 and b=(√2) we get ∫_1 ^(√2) (√(((√2) −x)(x−1))) dx=((((√2) −1)^2 )/8)=((3−2(√2))/8) .
$${let}\:{put}\:{I}=\:\int_{{a}} ^{{b}} \sqrt{\left({b}−{x}\right)\left({x}−{a}\right)}\:{dx}\:\:{the}\:{ch}.{x}=\frac{{a}−{b}}{\mathrm{2}}{t}\:+\:\frac{{a}+{b}}{\mathrm{2}} \\ $$$${b}−{x}={b}−\frac{{a}+{b}}{\mathrm{2}}\:−\frac{{a}−{b}}{\mathrm{2}}{t}=\frac{{b}−{a}}{\mathrm{2}}\:−\frac{{a}−{b}}{\mathrm{2}}{t}=\frac{{b}−{a}}{\mathrm{2}}\left(\mathrm{1}+{t}\right) \\ $$$${x}−{a}=\frac{{a}−{b}}{\mathrm{2}}{t}\:+\frac{{a}+{b}}{\mathrm{2}}\:−{a}=\:\frac{{a}−{b}}{\mathrm{2}}{t}\:+\frac{{b}−{a}}{\mathrm{2}}=\frac{{b}−{a}}{\mathrm{2}}\left(\mathrm{1}−{t}\right)\:{and} \\ $$$${we}\:{have}\:\frac{{a}−{b}}{\mathrm{2}}{t}={x}−\frac{{a}+{b}}{\mathrm{2}}\:\Rightarrow{t}=\frac{\mathrm{2}\left({x}−\frac{{a}+{b}}{\mathrm{2}}\right)}{{a}−{b}}=\frac{\mathrm{2}{x}\:−{a}−{b}}{{a}−{b}} \\ $$$${x}={a}\:\Rightarrow{t}=\mathrm{1}\:{and}\:{x}={b}\:\Rightarrow{t}=−\mathrm{1}\:\Rightarrow \\ $$$${I}\:=\:−\int_{−\mathrm{1}} ^{\mathrm{1}\:} \sqrt{\frac{{b}−{a}}{\mathrm{2}}\left(\mathrm{1}+{t}\right)\frac{{b}−{a}}{\mathrm{2}}\left(\mathrm{1}−{t}\right)}\:\frac{{a}−{b}}{\mathrm{2}}{dt} \\ $$$${I}=\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{4}}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} \:}\:{dt}=\:\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\:{ch}.{t}={sin}\theta \\ $$$${give}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} \:}\:{dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta\:{cos}\theta{d}\theta=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right){d}\theta\:=\:\frac{\pi}{\mathrm{4}}\:\Rightarrow\:{I}=\:\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{8}}\:.{then}\:{let}\:{take}\right. \\ $$$${a}=\mathrm{1}\:{and}\:{b}=\sqrt{\mathrm{2}}\:{we}\:{get} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \sqrt{\left(\sqrt{\mathrm{2}}\:−{x}\right)\left({x}−\mathrm{1}\right)}\:{dx}=\frac{\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{8}}=\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}}\:. \\ $$$$ \\ $$

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