What is the integral from 0 to 2 of e^{|1-x|} ?

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

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integral from 0 to 2 of e^{|1-x|}

\int_{0}^{2} e^{∣ 1 - x ∣} d x

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

Decimal

3.43656 \dots

Solution steps

\int_{0}^{2} e^{∣ 1 - x ∣} d x

Eliminate Absolutes

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

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

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

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

\int_{0}^{1} \frac{1}{x ^{2} + x + 1} d x

\int_{0}^{4} x^{2} - x d x

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

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

\int_{0}^{1} y sin (1) d y

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