What is the integral from 0 to t of e^{-2t} ?

The integral from 0 to t of e^{-2t} is -1/2 (e^{-2t}-1)

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integral from 0 to t of e^{-2t}

\int_{0}^{t} e^{- 2 t} d t

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

Solution steps

\int_{0}^{t} e^{- 2 t} d t

Apply u-substitution

: \int_{0}^{- 2 t} - \frac{1}{2} e^{u} d u

= \int_{0}^{- 2 t} - \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

= - \frac{1}{2} \cdot \int_{0}^{- 2 t} e^{u} d u

Use the common integral:

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

= - \frac{1}{2} [e^{u}]_{0}^{- 2 t}

Compute the boundaries:

e^{- 2 t} - 1

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

\int_{- 1}^{0} x e^{- x} d x

\int_{e}^{e^{3}} \frac{ln ( x )}{x} d x

\int_{2}^{2} 1

\int_{x^{2}}^{x} \frac{x}{y} d y

\int_{1}^{2} \frac{x ^{2} + 5}{x} d x

What is the integral from 0 to t of e^{-2t} ?
The integral from 0 to t of e^{-2t} is -1/2 (e^{-2t}-1)