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

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

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

integral

(e^{- y})^{2}

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

Solution steps

\int (e^{- y})^{2} d y

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = e^{- y}

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

Simplify

- \frac{( e ^{- y} ) ^{2}}{2} : - \frac{1}{2} e^{- 2 y}

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

Add a constant to the solution

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

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

2 t + 1

\frac{x ^{2} - 1}{x}

1 + 4 x^{2}

\frac{1}{x ^{\frac{1}{3}}}

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