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

The integral from 0 to 1 of (t-t^2)e^t is -e+3

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

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

- e + 3

Decimal

0.28171 \dots

Solution steps

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

Expand

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

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

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

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

= 1 - (e - 2)

Simplify

= - e + 3

What is the integral from 0 to 1 of (t-t^2)e^t ?
The integral from 0 to 1 of (t-t^2)e^t is -e+3