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

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

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

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

\frac{3724}{3}

Decimal

1241.33333 \dots

Solution steps

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

Expand

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

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

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 t^{2} d t - \int_{0}^{4} 16 t d t + \int_{0}^{4} 1 d t

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

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

\int_{0}^{4} 1 d t = 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 (8t-1)^2 ?
The integral from 0 to 4 of (8t-1)^2 is 3724/3