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

Decimal Notation

0.71828 \dots

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

f (x) = \frac{ln ( x )}{13 + x}

\int_{1}^{12} 2 x d x

\int \frac{9 e ^{2 t}}{1 + e ^{4 t}} d t

f (x) = \frac{1}{2} x^{2} + 7 x

\int co tan (x) 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