For-a-lt-b-then-a-b-x-a-x-b-dx-equal-to-

Question Number 119696 by bemath last updated on 26/Oct/20

For a<b then ∫_a ^b (x−a)(x−b) dx equal to _

$${For}\:{a}<{b}\:{then}\:\underset{{a}} {\overset{{b}} {\int}}\:\left({x}−{a}\right)\left({x}−{b}\right)\:{dx}\: \\ $$$${equal}\:{to}\:\_ \\ $$

Answered by TANMAY PANACEA last updated on 26/Oct/20

∫_a ^b (x^2 −x(a+b)+ab) dx =∣(x^3 /3)−(a+b)(x^2 /2)+abx∣_a ^b =(1/3)(b^3 −a^3 )−(((b+a)(b^2 −a^2 ))/2)+ab(b−a) =((2b^3 −2a^3 −3(b^3 −a^2 b+ab^2 −a^3 )+6ab^2 −6a^2 b)/6) =((2b^3 −2a^3 −3b^3 +3a^2 b−3ab^2 +3a^3 +6ab^2 −6a^2 b)/6) =((−b^3 +a^3 −3a^2 b+3ab^2 )/6) =(((a−b)^3 )/6)

$$\int_{{a}} ^{{b}} \left({x}^{\mathrm{2}} −{x}\left({a}+{b}\right)+{ab}\right)\:{dx} \\ $$$$=\mid\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\left({a}+{b}\right)\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{abx}\mid_{{a}} ^{{b}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({b}^{\mathrm{3}} −{a}^{\mathrm{3}} \right)−\frac{\left({b}+{a}\right)\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\mathrm{2}}+{ab}\left({b}−{a}\right) \\ $$$$=\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{2}{a}^{\mathrm{3}} −\mathrm{3}\left({b}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}+{ab}^{\mathrm{2}} −{a}^{\mathrm{3}} \right)+\mathrm{6}{ab}^{\mathrm{2}} −\mathrm{6}{a}^{\mathrm{2}} {b}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{2}{a}^{\mathrm{3}} −\mathrm{3}{b}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {b}−\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{3}} +\mathrm{6}{ab}^{\mathrm{2}} −\mathrm{6}{a}^{\mathrm{2}} {b}}{\mathrm{6}} \\ $$$$=\frac{−{b}^{\mathrm{3}} +{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} }{\mathrm{6}}\: \\ $$$$=\frac{\left({a}−{b}\right)^{\mathrm{3}} }{\mathrm{6}} \\ $$$$ \\ $$

Commented by TANMAY PANACEA last updated on 26/Oct/20

$${most}\:{welcome}\:{sir} \\ $$

Commented by Ar Brandon last updated on 26/Oct/20

I found my mistake. Thanks Sir

Answered by mathmax by abdo last updated on 26/Oct/20

∫_a ^b (x−a)(x−b)dx =_(x−a=t) ∫_0 ^(b−a) t(t+a−b)dt =∫_0 ^(b−a) (t^2 +(a−b)t)dt =[(t^3 /3) +((a−b)/2)t^2 ]_0 ^(b−a) =(((b−a)^3 )/3) +((a−b)/2)(b−a)^2 =(((b−a)^3 )/3)−(((b−a)^3 )/2) =−(((b−a)^3 )/6)

$$\int_{\mathrm{a}} ^{\mathrm{b}} \left(\mathrm{x}−\mathrm{a}\right)\left(\mathrm{x}−\mathrm{b}\right)\mathrm{dx}\:=_{\mathrm{x}−\mathrm{a}=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{b}−\mathrm{a}} \mathrm{t}\left(\mathrm{t}+\mathrm{a}−\mathrm{b}\right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{b}−\mathrm{a}} \left(\mathrm{t}^{\mathrm{2}} +\left(\mathrm{a}−\mathrm{b}\right)\mathrm{t}\right)\mathrm{dt}\:=\left[\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{3}}\:+\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{b}−\mathrm{a}} \\ $$$$=\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{3}} }{\mathrm{3}}\:+\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} \:=\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{3}} }{\mathrm{3}}−\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{3}} }{\mathrm{2}} \\ $$$$=−\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{3}} }{\mathrm{6}} \\ $$

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