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

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

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

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

Expand

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

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 2 x d x = 4

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

= \frac{8}{3} - 4 + 2

Simplify

\frac{8}{3} - 4 + 2 : \frac{2}{3}

= \frac{2}{3}

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

\frac{d}{d x} ((x^{3} + 4)^{- 3})

x \to 0 + lim (6 - \frac{1}{x ^{2}})

y = 4 x^{\frac{2}{3}} + 6 x^{\frac{1}{2}} - 8

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