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

The integral from 0 to 1 of (e^t-e+1)^2 is -e^2+2e+(e^2-1)/2-1

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

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

- e^{2} + 2 e + \frac{e ^{2} - 1}{2} - 1

Solution steps

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

Expand

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

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

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

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

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

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

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

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

= e^{2} - 2 e - (2 e^{2} - 2 e) + \frac{e ^{2} - 1}{2} + 2 (e - 1) + 1

Simplify

e^{2} - 2 e - (2 e^{2} - 2 e) + \frac{e ^{2} - 1}{2} + 2 (e - 1) + 1 : - e^{2} + 2 e + \frac{e ^{2} - 1}{2} - 1

= - e^{2} + 2 e + \frac{e ^{2} - 1}{2} - 1

\int_{0}^{1} 1

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

\int_{2}^{5} 2 π y (5 - y) d y

\int_{0}^{4} \frac{x ^{2}}{16} d x

\int_{1}^{2} - \frac{1}{x} d x

What is the integral from 0 to 1 of (e^t-e+1)^2 ?
The integral from 0 to 1 of (e^t-e+1)^2 is -e^2+2e+(e^2-1)/2-1