Question Number 80752 by bagjamath last updated on 06/Feb/20
Commented by john santu last updated on 06/Feb/20
$${x}−\mathrm{2010}=\:{t}\: \\ $$$${x}−\mathrm{2009}={t}+\mathrm{1} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{0}} \left({t}^{\mathrm{2}} +{t}\right){dt}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \:\mid_{−\mathrm{1}} ^{\mathrm{0}} \\ $$$$=\:−\left(−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Answered by MJS last updated on 06/Feb/20
![∫_a ^b (x−a)(x−b)dx= =∫_a ^b (x^2 −(a+b)x+ab)dx= =[(1/3)x^3 −((a+b)/2)x^2 +abx]_a ^b = =(1/6)(a^3 −b^3 )−(1/2)ab(a−b)= [b=a+c] =(1/6)(a^3 −(a+c)^3 )+(1/2)a(a+c)c=−(c^3 /6)= [c=1] =−(1/6)](https://www.tinkutara.com/question/Q80757.png)
$$\underset{{a}} {\overset{{b}} {\int}}\left({x}−{a}\right)\left({x}−{b}\right){dx}= \\ $$$$=\underset{{a}} {\overset{{b}} {\int}}\left({x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}\right){dx}= \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} −\frac{{a}+{b}}{\mathrm{2}}{x}^{\mathrm{2}} +{abx}\right]_{{a}} ^{{b}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)−\frac{\mathrm{1}}{\mathrm{2}}{ab}\left({a}−{b}\right)= \\ $$$$\:\:\:\:\:\left[{b}={a}+{c}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left({a}^{\mathrm{3}} −\left({a}+{c}\right)^{\mathrm{3}} \right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({a}+{c}\right){c}=−\frac{{c}^{\mathrm{3}} }{\mathrm{6}}= \\ $$$$\:\:\:\:\:\left[{c}=\mathrm{1}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by bagjamath last updated on 06/Feb/20
$${thank}\:{you}\:{Sir} \\ $$