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

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

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integral of (e^{1-x})

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

- e^{1 - x} + C

Solution steps

\int e^{1 - x} d x

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = 1 - x

= - e^{1 - x}

Add a constant to the solution

= - e^{1 - x} + C

\frac{d}{d x} (\frac{1}{10} x - \frac{1}{8})

n = 0 \sum \infty \frac{1}{n ^{- 1}}

\int (x^{- \frac{1}{3}}) d x

\int sin (b x) d x

\frac{1}{( s - 3 ) ^{2}}