What is the integral of-1/4 (x-1)(x-2)(x-4) ?

The integral of-1/4 (x-1)(x-2)(x-4) is -1/4 ((x^4)/4-(7x^3)/3+7x^2-8x)+C

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integral of-1/4 (x-1)(x-2)(x-4)

\int - \frac{1}{4} (x - 1) (x - 2) (x - 4) d x

- \frac{1}{4} (\frac{x ^{4}}{4} - \frac{7 x ^{3}}{3} + 7 x^{2} - 8 x) + C

Solution steps

\int - \frac{1}{4} (x - 1) (x - 2) (x - 4) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int (x - 1) (x - 2) (x - 4) d x

Expand

(x - 1) (x - 2) (x - 4) : x^{3} - 7 x^{2} + 14 x - 8

= - \frac{1}{4} \cdot \int x^{3} - 7 x^{2} + 14 x - 8 d x

Apply the Sum Rule:

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

= - \frac{1}{4} (\int x^{3} d x - \int 7 x^{2} d x + \int 14 x d x - \int 8 d x)

\int x^{3} d x = \frac{x ^{4}}{4}

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 14 x d x = 7 x^{2}

\int 8 d x = 8 x

= - \frac{1}{4} (\frac{x ^{4}}{4} - \frac{7 x ^{3}}{3} + 7 x^{2} - 8 x)

Add a constant to the solution

= - \frac{1}{4} (\frac{x ^{4}}{4} - \frac{7 x ^{3}}{3} + 7 x^{2} - 8 x) + C

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What is the integral of-1/4 (x-1)(x-2)(x-4) ?
The integral of-1/4 (x-1)(x-2)(x-4) is -1/4 ((x^4)/4-(7x^3)/3+7x^2-8x)+C