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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int_{0}^{- 2 x} - \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 x} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e^{- 2 x} - 1

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

\int_{0}^{3} \frac{1}{x ^{2} - x + \frac{5}{2}} d x

\int_{2}^{5} x e^{2 x} d x

\int_{5}^{2} 1

\int_{- 2}^{- 1} \frac{x - 1}{2 x ^{2} + 8 x} d x

\int_{0}^{\frac{9}{8}} \frac{1}{64 x ^{2} + 81} d x

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