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

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

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

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

Expand

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

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

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

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

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

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

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

Simplify

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

= \frac{2}{3}

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