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

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

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integral from 1 to-1 of (x-1)(3x+2)

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

2

Solution steps

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

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

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

Expand

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

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

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

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

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

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

= - (2 - 0 - 4)

Simplify

= 2

\frac{d}{d x} (\frac{2}{sin ( x ) + cos ( x )})

t \frac{d y}{d t} = t^{3} + 25 t^{3} y

\int (y sin (x y)) d x

\frac{\partial}{\partial x} (y^{2} - 4 x^{2} + 7 y + 5 x + 13)

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