What is the integral of 1/(e^{-2y)} ?

The integral of 1/(e^{-2y)} is 1/2 e^{2y}+C

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integral of 1/(e^{-2y)}

\int \frac{1}{e ^{- 2 y}} d y

\frac{1}{2} e^{2 y} + C

Solution steps

\int \frac{1}{e ^{- 2 y}} d y

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{- 2 y}} = e^{2 y}

= \int e^{2 y} d y

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = 2 y

= \frac{1}{2} e^{2 y}

Add a constant to the solution

= \frac{1}{2} e^{2 y} + C

\int_{0}^{\frac{1}{2}} 30 x d x

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\frac{d}{d t} (- 1 cos (t) + 8 sin (t))

What is the integral of 1/(e^{-2y)} ?
The integral of 1/(e^{-2y)} is 1/2 e^{2y}+C