Question Number 164 by 123456 last updated on 25/Jan/15
$$\mathrm{evaluate}\:{f}\left({a},{b}\right)=\underset{{a}} {\overset{{b}} {\int}}\left({b}−{x}\right)\left({x}−{a}\right){dx} \\ $$
Answered by prakash jain last updated on 14/Dec/14
![(b−x)(x−a)=−x^2 +(a+b)x−ab ∫_a ^b (−x^2 +(a+b)x−ab)dx =[((−x^3 )/3)+(((a+b)x^2 )/2)−abx]_a ^b =[((−b^3 )/3)+(((a+b)b^2 )/2)−ab^2 ]−[((−a^3 )/3)+(((a+b)a^2 )/2)−a^2 b] =[(b^3 /6)−((ab^2 )/2)]−[(a^3 /6)−((a^2 b)/2)] =((b^3 −a^3 −3ab^2 +3a^2 b)/6)=(((b−a)^3 )/6) f(a,b)=(((b−a)^3 )/6) So f(a,b)≠f(b,a)](https://www.tinkutara.com/question/Q175.png)
$$\left({b}−{x}\right)\left({x}−{a}\right)=−{x}^{\mathrm{2}} +\left({a}+{b}\right){x}−{ab} \\ $$$$\int_{{a}} ^{{b}} \left(−{x}^{\mathrm{2}} +\left({a}+{b}\right){x}−{ab}\right){dx} \\ $$$$=\left[\frac{−{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({a}+{b}\right){x}^{\mathrm{2}} }{\mathrm{2}}−{abx}\right]_{{a}} ^{{b}} \\ $$$$=\left[\frac{−{b}^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({a}+{b}\right){b}^{\mathrm{2}} }{\mathrm{2}}−{ab}^{\mathrm{2}} \right]−\left[\frac{−{a}^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({a}+{b}\right){a}^{\mathrm{2}} }{\mathrm{2}}−{a}^{\mathrm{2}} {b}\right] \\ $$$$=\left[\frac{{b}^{\mathrm{3}} }{\mathrm{6}}−\frac{{ab}^{\mathrm{2}} }{\mathrm{2}}\right]−\left[\frac{{a}^{\mathrm{3}} }{\mathrm{6}}−\frac{{a}^{\mathrm{2}} {b}}{\mathrm{2}}\right] \\ $$$$=\frac{{b}^{\mathrm{3}} −{a}^{\mathrm{3}} −\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} {b}}{\mathrm{6}}=\frac{\left({b}−{a}\right)^{\mathrm{3}} }{\mathrm{6}} \\ $$$${f}\left({a},{b}\right)=\frac{\left({b}−{a}\right)^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\mathrm{S}{o}\:{f}\left({a},{b}\right)\neq{f}\left({b},{a}\right) \\ $$