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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

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

Expand

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

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

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 4 x d x = 16

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

= \frac{26}{3} - 16 + 8

Simplify

\frac{26}{3} - 16 + 8 : \frac{2}{3}

= \frac{2}{3}

\int_{- 1}^{2} x^{2} - 4 x + 3 d x

\int_{- y}^{y} \frac{1}{4} (y - x) e^{- y} d x

\int_{- π}^{π} sin^{2} (3 x) d x

\int_{π}^{π} x sin (x) d x

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

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