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

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

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

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

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

Solution steps

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

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_{- 1}^{4} - x^{2} + 3 x + 4 d x

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\int_{\frac{π}{6}}^{\frac{5 π}{6}} sin (x) d x

\int_{- 1}^{1} \frac{1}{∣ x ∣} d x

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