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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{1}{s s ^{2} - 9} d s

\int (- 4 x^{6} + 2 x^{5} - 3 x^{3} + 3) d x

\int (3 x^{3} + 4 x - 3) d x

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

\int \frac{x ( x + 2 )}{x ^{3} + 3 x ^{2} + 1} d x