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

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

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

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

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

Decimal

- 1

Solution steps

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

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

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

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

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