What is the integral from 0 to 1 of 2t(t+2) ?

The integral from 0 to 1 of 2t(t+2) is 8/3

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integral from 0 to 1 of 2t(t+2)

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

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{1} t (t + 2) d t

Expand

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

= 2 \cdot \int_{0}^{1} t^{2} + 2 t 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

= 2 (\int_{0}^{1} t^{2} d t + \int_{0}^{1} 2 t d t)

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

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

= 2 (\frac{1}{3} + 1)

Simplify

2 (\frac{1}{3} + 1) : \frac{8}{3}

= \frac{8}{3}

What is the integral from 0 to 1 of 2t(t+2) ?
The integral from 0 to 1 of 2t(t+2) is 8/3