What is the integral from 1 to 3 of (x-1)(3-x) ?

The integral from 1 to 3 of (x-1)(3-x) is 4/3

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integral from 1 to 3 of (x-1)(3-x)

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Expand

(x - 1) (3 - x) : 4 x - x^{2} - 3

= \int_{1}^{3} 4 x - x^{2} - 3 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

= \int_{1}^{3} 4 x d x - \int_{1}^{3} x^{2} d x - \int_{1}^{3} 3 d x

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 3 d x = 6

= 16 - \frac{26}{3} - 6

Simplify

16 - \frac{26}{3} - 6 : \frac{4}{3}

= \frac{4}{3}

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\int_{- 2}^{6} - 2 x + 16 d x

What is the integral from 1 to 3 of (x-1)(3-x) ?
The integral from 1 to 3 of (x-1)(3-x) is 4/3