What is the integral from 1 to 2 of 6(x-1)(2-x) ?

The integral from 1 to 2 of 6(x-1)(2-x) is 1

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integral from 1 to 2 of 6(x-1)(2-x)

\int_{1}^{2} 6 (x - 1) (2 - x) d x

1

Solution steps

\int_{1}^{2} 6 (x - 1) (2 - x) d x

Take the constant out:

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

= 6 \cdot \int_{1}^{2} (x - 1) (2 - x) d x

Expand

(x - 1) (2 - x) : 3 x - x^{2} - 2

= 6 \cdot \int_{1}^{2} 3 x - x^{2} - 2 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

= 6 (\int_{1}^{2} 3 x d x - \int_{1}^{2} x^{2} d x - \int_{1}^{2} 2 d x)

\int_{1}^{2} 3 x d x = \frac{9}{2}

\int_{1}^{2} x^{2} d x = \frac{7}{3}

\int_{1}^{2} 2 d x = 2

= 6 (\frac{9}{2} - \frac{7}{3} - 2)

Simplify

6 (\frac{9}{2} - \frac{7}{3} - 2) : 1

= 1

\int_{0}^{2} (12 - x) x d x

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\int_{- x}^{x^{2} + 1} (x^{3} - x) d x

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What is the integral from 1 to 2 of 6(x-1)(2-x) ?
The integral from 1 to 2 of 6(x-1)(2-x) is 1