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

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

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

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

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

Solution steps

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

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 e^{- s t} (t - 2)^{2} d t

\int_{x}^{y} (t^{2} - 1) d t

f^{^{'}} (x) = x^{- 3} + π x^{2} - 2

f (2) = 10 - 3 x

x \to 2 + lim (2 x)

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