What is the integral from 0 to 16 of sqrt(t^2+2t+1) ?

The integral from 0 to 16 of sqrt(t^2+2t+1) is 144

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integral from 0 to 16 of sqrt(t^2+2t+1)

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

144

Solution steps

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

Factor

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

= \int_{0}^{16} (t + 1)^{2} d t

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{16} (t + 1)^{2} d t

Eliminate Absolutes

= \int_{0}^{16} (t + 1)^{2} d t

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{16} ∣ (t + 1) ∣ d t

Eliminate Absolutes

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

\int_{0}^{16} t d t = 128

\int_{0}^{16} 1 d t = 16

= 128 + 16

Simplify

= 144

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\frac{d}{d x} (- 9 x^{3} + 27 x + 7 x - 66)

What is the integral from 0 to 16 of sqrt(t^2+2t+1) ?
The integral from 0 to 16 of sqrt(t^2+2t+1) is 144