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

The integral of (e^{-2x})^2 is -1/4 e^{-4x}+C

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

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

- \frac{1}{4} e^{- 4 x} + C

Solution steps

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

Simplify

(e^{- 2 x})^{2} : e^{- 4 x}

= \int e^{- 4 x} d x

Apply u-substitution

: \int - \frac{1}{4} e^{u} d u

= \int - \frac{1}{4} e^{u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{4} e^{u}

Substitute back

u = - 4 x

= - \frac{1}{4} e^{- 4 x}

Add a constant to the solution

= - \frac{1}{4} e^{- 4 x} + C

x \to 2 lim (2 x + 2)

\frac{\partial}{\partial x} (e^{2 z})

f (x) = 3 x^{2} - 4, at x = 2

y^{^{''}} - y = e^{- t}, y (0) = 1, y^{^{'}} (0) = 0

\int 8 + 7^{x} d x

What is the integral of (e^{-2x})^2 ?
The integral of (e^{-2x})^2 is -1/4 e^{-4x}+C