What is the integral of-Be^{-2t} ?

The integral of-Be^{-2t} is 1/2 Be^{-2t}+C

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integral of-Be^{-2t}

\int - B e^{- 2 t} d t

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

Solution steps

\int - B e^{- 2 t} d t

Take the constant out:

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

= - B \cdot \int e^{- 2 t} d t

Apply u-substitution

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

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

Take the constant out:

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

= - B (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= - B (- \frac{1}{2} e^{u})

Substitute back

u = - 2 t

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

Simplify

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

Add a constant to the solution

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

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