What is the integral from x to 0 of e^{-16t} ?

The integral from x to 0 of e^{-16t} is -1/16 (1-e^{-16x})

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integral from x to 0 of e^{-16t}

\int_{x}^{0} e^{- 16 t} d t

- \frac{1}{16} (1 - e^{- 16 x})

Solution steps

\int_{x}^{0} e^{- 16 t} d t

Apply u-substitution

: \int_{- 16 x}^{0} - \frac{1}{16} e^{u} d u

= \int_{- 16 x}^{0} - \frac{1}{16} e^{u} d u

Take the constant out:

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

= - \frac{1}{16} \cdot \int_{- 16 x}^{0} e^{u} d u

Use the common integral:

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

= - \frac{1}{16} [e^{u}]_{- 16 x}^{0}

Compute the boundaries:

1 - e^{- 16 x}

= - \frac{1}{16} (1 - e^{- 16 x})

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What is the integral from x to 0 of e^{-16t} ?
The integral from x to 0 of e^{-16t} is -1/16 (1-e^{-16x})