What is the integral from 1 to e of 1+e-x-ln(x) ?

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

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integral from 1 to e of 1+e-x-ln(x)

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

e^{2} - 2 - \frac{e ^{2} - 1}{2}

Decimal Notation

2.19452 \dots

Solution steps

\int_{1}^{e} 1 + e - x - ln (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_{1}^{e} 1 d x + \int_{1}^{e} e d x - \int_{1}^{e} x d x - \int_{1}^{e} ln (x) d x

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

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

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

\int_{1}^{e} ln (x) d x = 1

= e - 1 + e^{2} - e - \frac{e ^{2} - 1}{2} - 1

Simplify

= e^{2} - 2 - \frac{e ^{2} - 1}{2}

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