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

The integral from 0 to 2 of (4t-1)^2 is 86/3

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

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

\frac{86}{3}

Decimal

28.66666 \dots

Solution steps

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

Expand

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

= \int_{0}^{2} 16 t^{2} - 8 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}^{2} 16 t^{2} d t - \int_{0}^{2} 8 t d t + \int_{0}^{2} 1 d t

\int_{0}^{2} 16 t^{2} d t = \frac{128}{3}

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

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

= \frac{128}{3} - 16 + 2

Simplify

\frac{128}{3} - 16 + 2 : \frac{86}{3}

= \frac{86}{3}

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What is the integral from 0 to 2 of (4t-1)^2 ?
The integral from 0 to 2 of (4t-1)^2 is 86/3