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

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

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

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

e - 2

Solution steps

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

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

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

= e - 1 - 1

Simplify

= e - 2

\int_{0}^{0.6} 10000 x d x

9 x^{3}

(x \cdot x - 1) y^{^{'}} - x y (1 - y) = 0

\frac{d}{d t} (\frac{3 t ^{2}}{1 + 4 t ^{2} + 9 t ^{4}})

\int e^{x^{2}} x^{3} d x

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