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

F o r a < b t h e n \underset{a}{\int^{b}} (x - a) (x - b) d x

e q u a l t o_

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

\int_{a}^{b} (x^{2} - x (a + b) + a b) d x

=∣ \frac{x^{3}}{3} - (a + b) \frac{x^{2}}{2} + a b x ∣_{a}^{b}

= \frac{1}{3} (b^{3} - a^{3}) - \frac{(b + a) (b^{2} - a^{2})}{2} + a b (b - a)

= \frac{2 b^{3} - 2 a^{3} - 3 (b^{3} - a^{2} b + {a b}^{2} - a^{3}) + 6 {a b}^{2} - 6 a^{2} b}{6}

= \frac{2 b^{3} - 2 a^{3} - 3 b^{3} + 3 a^{2} b - 3 {a b}^{2} + 3 a^{3} + 6 {a b}^{2} - 6 a^{2} b}{6}

= \frac{- b^{3} + a^{3} - 3 a^{2} b + 3 {a b}^{2}}{6}

= \frac{{(a - b)}^{3}}{6}

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

m o s t w e l c o m e s i r

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

\int_{a}^{b} (x - a) (x - b) dx =_{x - a = t} \int_{0}^{b - a} t (t + a - b) dt

= \int_{0}^{b - a} (t^{2} + (a - b) t) dt = {[\frac{t^{3}}{3} + \frac{a - b}{2} t^{2}]}_{0}^{b - a}

= \frac{{(b - a)}^{3}}{3} + \frac{a - b}{2} {(b - a)}^{2} = \frac{{(b - a)}^{3}}{3} - \frac{{(b - a)}^{3}}{2}

= - \frac{{(b - a)}^{3}}{6}