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

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

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

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

e - 1

Decimal Notation

1.71828 \dots

Solution steps

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

Expand

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

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

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

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

= e - 2 + 1

Simplify

= e - 1

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