What is the integral from 0 to 2 of (x-1)(2x+1) ?

The integral from 0 to 2 of (x-1)(2x+1) is 4/3

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integral from 0 to 2 of (x-1)(2x+1)

\int_{0}^{2} (x - 1) (2 x + 1) d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{0}^{2} (x - 1) (2 x + 1) d x

Expand

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

= \int_{0}^{2} 2 x^{2} - x - 1 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}^{2} 2 x^{2} d x - \int_{0}^{2} x d x - \int_{0}^{2} 1 d x

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

\int_{0}^{2} x d x = 2

\int_{0}^{2} 1 d x = 2

= \frac{16}{3} - 2 - 2

Simplify

\frac{16}{3} - 2 - 2 : \frac{4}{3}

= \frac{4}{3}

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What is the integral from 0 to 2 of (x-1)(2x+1) ?
The integral from 0 to 2 of (x-1)(2x+1) is 4/3