What is the integral of e^{(-y)/2} ?

The integral of e^{(-y)/2} is -2e^{-y/2}+C

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integral of e^{(-y)/2}

integral

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

- 2 e^{- \frac{y}{2}} + C

Solution steps

\int e^{\frac{- y}{2}} d y

Simplify

= \int e^{- \frac{y}{2}} d y

Apply u-substitution

: \int - 2 e^{u} d u

= \int - 2 e^{u} d u

Take the constant out:

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

= - 2 \cdot \int e^{u} d u

Use the common integral:

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

= - 2 e^{u}

Substitute back

u = - \frac{y}{2}

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

Add a constant to the solution

= - 2 e^{- \frac{y}{2}} + C

\frac{e ^{2} x}{2}

1 - tan^{2} (x)

\frac{x ^{3}}{( x + 2 ) ^{4}}

x^{3} (x^{2} + 4)^{\frac{1}{2}}

\frac{x ^{4}}{( x ^{3} - 1 )}