What is the integral from 0 to 2 of (8x-1)^2 ?

The integral from 0 to 2 of (8x-1)^2 is 422/3

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integral from 0 to 2 of (8x-1)^2

\int_{0}^{2} (8 x - 1)^{2} d x

\frac{422}{3}

Decimal

140.66666 \dots

Solution steps

\int_{0}^{2} (8 x - 1)^{2} d x

Expand

(8 x - 1)^{2} : 64 x^{2} - 16 x + 1

= \int_{0}^{2} 64 x^{2} - 16 x + 1 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} 64 x^{2} d x - \int_{0}^{2} 16 x d x + \int_{0}^{2} 1 d x

\int_{0}^{2} 64 x^{2} d x = \frac{512}{3}

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

\int_{0}^{2} 1 d x = 2

= \frac{512}{3} - 32 + 2

Simplify

\frac{512}{3} - 32 + 2 : \frac{422}{3}

= \frac{422}{3}

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What is the integral from 0 to 2 of (8x-1)^2 ?
The integral from 0 to 2 of (8x-1)^2 is 422/3