What is the integral from 0 to 4 of (8x-1)^2 ?

The integral from 0 to 4 of (8x-1)^2 is 3724/3

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integral from 0 to 4 of (8x-1)^2

\int_{0}^{4} (8 x - 1)^{2} d x

\frac{3724}{3}

Decimal

1241.33333 \dots

Solution steps

\int_{0}^{4} (8 x - 1)^{2} d x

Expand

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

= \int_{0}^{4} 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}^{4} 64 x^{2} d x - \int_{0}^{4} 16 x d x + \int_{0}^{4} 1 d x

\int_{0}^{4} 64 x^{2} d x = \frac{4096}{3}

\int_{0}^{4} 16 x d x = 128

\int_{0}^{4} 1 d x = 4

= \frac{4096}{3} - 128 + 4

Simplify

\frac{4096}{3} - 128 + 4 : \frac{3724}{3}

= \frac{3724}{3}

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What is the integral from 0 to 4 of (8x-1)^2 ?
The integral from 0 to 4 of (8x-1)^2 is 3724/3