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

The integral from 0 to 2 of (x-4)^2 is 56/3

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

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

\frac{56}{3}

Decimal

18.66666 \dots

Solution steps

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

Expand

(x - 4)^{2} : x^{2} - 8 x + 16

= \int_{0}^{2} x^{2} - 8 x + 16 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} 8 x d x + \int_{0}^{2} 16 d x

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

\int_{0}^{2} 8 x d x = 16

\int_{0}^{2} 16 d x = 32

= \frac{8}{3} - 16 + 32

Simplify

\frac{8}{3} - 16 + 32 : \frac{56}{3}

= \frac{56}{3}

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

\int_{0}^{1} (\frac{6}{x ^{2} + 1}) d x

\int_{0}^{1} ∣ x (2 x - 1) ∣ d x

\int_{0}^{3} e^{- t^{2}} d t

\int_{1}^{e} \frac{( 2 x ^{2} + 1 )}{2 x} d x

What is the integral from 0 to 2 of (x-4)^2 ?
The integral from 0 to 2 of (x-4)^2 is 56/3