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

The integral from 0 to 4 of (2t-1)^2 is 172/3

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

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

\frac{172}{3}

Decimal

57.33333 \dots

Solution steps

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

Expand

(2 t - 1)^{2} : 4 t^{2} - 4 t + 1

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

\int_{0}^{4} 4 t^{2} d t = \frac{256}{3}

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

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

= \frac{256}{3} - 32 + 4

Simplify

\frac{256}{3} - 32 + 4 : \frac{172}{3}

= \frac{172}{3}

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What is the integral from 0 to 4 of (2t-1)^2 ?
The integral from 0 to 4 of (2t-1)^2 is 172/3