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

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

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

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= \frac{16}{3}

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\int_{s i n (x)}^{1} 1 + x^{2} d x

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

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