What is the integral from 0 to 4 of (y-2)^2 ?

The integral from 0 to 4 of (y-2)^2 is 16/3

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integral from 0 to 4 of (y-2)^2

\int_{0}^{4} (y - 2)^{2} d y

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

\int_{0}^{4} (y - 2)^{2} d y

Expand

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

= \int_{0}^{4} 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_{0}^{4} y^{2} d y - \int_{0}^{4} 4 y d y + \int_{0}^{4} 4 d y

\int_{0}^{4} y^{2} d y = \frac{64}{3}

\int_{0}^{4} 4 y d y = 32

\int_{0}^{4} 4 d y = 16

= \frac{64}{3} - 32 + 16

Simplify

\frac{64}{3} - 32 + 16 : \frac{16}{3}

= \frac{16}{3}

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

\int_{0}^{3} (- x^{2} + 4) d x

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

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

\int_{- 9}^{0} (\frac{y}{9} + y + 10) d y

What is the integral from 0 to 4 of (y-2)^2 ?
The integral from 0 to 4 of (y-2)^2 is 16/3