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

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

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

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

- e + 3

Solution steps

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

Expand

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

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

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^{x} x d x - \int_{0}^{1} e^{x} x^{2} d x

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

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

= 1 - (e - 2)

Simplify

= - e + 3

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