What is the integral from a to b of (x-a)(x-b) ?

The integral from a to b of (x-a)(x-b) is ((-a+b)^3}{12}-a^2b+ab^2-\frac{(b-a)(-a-b)^2)/4

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integral from a to b of (x-a)(x-b)

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

\frac{( - a + b ) ^{3}}{12} - a^{2} b + a b^{2} - \frac{( b - a ) ( - a - b ) ^{2}}{4}

Solution steps

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

Complete the square

(x - a) (x - b) : (x + \frac{- b - a}{2})^{2} + ab - \frac{( - b - a ) ^{2}}{4}

= \int_{a}^{b} (x + \frac{- b - a}{2})^{2} + ab - \frac{( - b - a ) ^{2}}{4} d x

Apply u-substitution

: \int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} u^{2} + ab - \frac{( - b - a ) ^{2}}{4} d u

= \int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} u^{2} + ab - \frac{( - b - a ) ^{2}}{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} u^{2} d u + \int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} ab d u - \int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} \frac{( - b - a ) ^{2}}{4} d u

\int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} u^{2} d u = \frac{( - a + b ) ^{3}}{12}

\int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} ab d u = - a^{2} b + a b^{2}

\int_{a + \frac{- b - a}{2}}^{b + \frac{- b - a}{2}} \frac{( - b - a ) ^{2}}{4} d u = \frac{( b - a ) ( - a - b ) ^{2}}{4}

= \frac{( - a + b ) ^{3}}{12} - a^{2} b + a b^{2} - \frac{( b - a ) ( - a - b ) ^{2}}{4}

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\int_{0}^{1} cos (3 x - 3) d x

What is the integral from a to b of (x-a)(x-b) ?
The integral from a to b of (x-a)(x-b) is ((-a+b)^3}{12}-a^2b+ab^2-\frac{(b-a)(-a-b)^2)/4