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

The integral from 0 to 1 of (4t-1)^2 is 7/3

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

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

\frac{7}{3}

Decimal

2.33333 \dots

Solution steps

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

Expand

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

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

\int_{0}^{1} 16 t^{2} d t = \frac{16}{3}

\int_{0}^{1} 8 t d t = 4

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

= \frac{16}{3} - 4 + 1

Simplify

\frac{16}{3} - 4 + 1 : \frac{7}{3}

= \frac{7}{3}

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What is the integral from 0 to 1 of (4t-1)^2 ?
The integral from 0 to 1 of (4t-1)^2 is 7/3