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

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

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integral from x to 1 of e^{(1-x)}

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

- 1 + e^{1 - x}

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int e^{u} d u = e^{u}

= - [e^{u}]_{1 - x}^{0}

Compute the boundaries:

1 - e^{1 - x}

= - (1 - e^{1 - x})

Simplify

= - 1 + e^{1 - x}

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