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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

= \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}

Substitute back

u = e^{- t}

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

Simplify

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

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

Add a constant to the solution

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

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