What is the integral from 2 to 6 of (1-t)^2 ?

The integral from 2 to 6 of (1-t)^2 is 124/3

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integral from 2 to 6 of (1-t)^2

\int_{2}^{6} (1 - t)^{2} d t

\frac{124}{3}

Decimal Notation

41.33333 \dots

Solution steps

\int_{2}^{6} (1 - t)^{2} d t

Expand

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

= \int_{2}^{6} 1 - 2 t + t^{2} 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_{2}^{6} 1 d t - \int_{2}^{6} 2 t d t + \int_{2}^{6} t^{2} d t

\int_{2}^{6} 1 d t = 4

\int_{2}^{6} 2 t d t = 32

\int_{2}^{6} t^{2} d t = \frac{208}{3}

= 4 - 32 + \frac{208}{3}

Simplify

4 - 32 + \frac{208}{3} : \frac{124}{3}

= \frac{124}{3}

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What is the integral from 2 to 6 of (1-t)^2 ?
The integral from 2 to 6 of (1-t)^2 is 124/3