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

The integral from-1 to 1 of (2x-1)^2 is 14/3

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

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

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

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

Expand

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

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

\int_{- 1}^{1} 4 x^{2} d x = \frac{8}{3}

\int_{- 1}^{1} 4 x d x = 0

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

= \frac{8}{3} - 0 + 2

Simplify

\frac{8}{3} - 0 + 2 : \frac{14}{3}

= \frac{14}{3}

\frac{d y}{d x} = - 5 - 5 x + y

\int 2 y d y

\frac{\partial}{\partial x} ((x - 5)^{\frac{1}{4}} y^{\frac{1}{4}})

d er i v a t i v e f (x) = e^{x} sin (x)

x \to 1 lim (x^{2} + 3)

What is the integral from-1 to 1 of (2x-1)^2 ?
The integral from-1 to 1 of (2x-1)^2 is 14/3