What is the integral from 0 to 18 of sqrt(t^2+2t+1) ?

The integral from 0 to 18 of sqrt(t^2+2t+1) is 180

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integral from 0 to 18 of sqrt(t^2+2t+1)

\int_{0}^{18} t^{2} + 2 t + 1 d t

180

Solution steps

\int_{0}^{18} t^{2} + 2 t + 1 d t

Factor

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

= \int_{0}^{18} (t + 1)^{2} d t

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{18} (t + 1)^{2} d t

Eliminate Absolutes

= \int_{0}^{18} (t + 1)^{2} d t

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{18} ∣ (t + 1) ∣ d t

Eliminate Absolutes

= \int_{0}^{18} 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}^{18} t d t + \int_{0}^{18} 1 d t

\int_{0}^{18} t d t = 162

\int_{0}^{18} 1 d t = 18

= 162 + 18

Simplify

= 180

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What is the integral from 0 to 18 of sqrt(t^2+2t+1) ?
The integral from 0 to 18 of sqrt(t^2+2t+1) is 180