What is the integral from 0 to 4 of (x-3)(x-1) ?

The integral from 0 to 4 of (x-3)(x-1) is 4/3

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integral from 0 to 4 of (x-3)(x-1)

\int_{0}^{4} (x - 3) (x - 1) d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{0}^{4} (x - 3) (x - 1) d x

Expand

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

= \int_{0}^{4} x^{2} - 4 x + 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_{0}^{4} x^{2} d x - \int_{0}^{4} 4 x d x + \int_{0}^{4} 3 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 4 x d x = 32

\int_{0}^{4} 3 d x = 12

= \frac{64}{3} - 32 + 12

Simplify

\frac{64}{3} - 32 + 12 : \frac{4}{3}

= \frac{4}{3}

What is the integral from 0 to 4 of (x-3)(x-1) ?
The integral from 0 to 4 of (x-3)(x-1) is 4/3