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

The integral from 2 to-1 of (2x+1)^2 is -21

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

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

- 21

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Expand

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

= - \int_{- 1}^{2} 4 x^{2} + 4 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_{- 1}^{2} 4 x^{2} d x + \int_{- 1}^{2} 4 x d x + \int_{- 1}^{2} 1 d x)

\int_{- 1}^{2} 4 x^{2} d x = 12

\int_{- 1}^{2} 4 x d x = 6

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

= - (12 + 6 + 3)

Simplify

= - 21

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What is the integral from 2 to-1 of (2x+1)^2 ?
The integral from 2 to-1 of (2x+1)^2 is -21