What is the integral from 0 to x^3 of e^{-5t} ?

The integral from 0 to x^3 of e^{-5t} is -1/5 (e^{-5x^3}-1)

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integral from 0 to x^3 of e^{-5t}

\int_{0}^{x^{3}} e^{- 5 t} d t

- \frac{1}{5} (e^{- 5 x^{3}} - 1)

Solution steps

\int_{0}^{x^{3}} e^{- 5 t} d t

Apply u-substitution

: \int_{0}^{- 5 x^{3}} - \frac{1}{5} e^{u} d u

= \int_{0}^{- 5 x^{3}} - \frac{1}{5} e^{u} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int_{0}^{- 5 x^{3}} e^{u} d u

Use the common integral:

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

= - \frac{1}{5} [e^{u}]_{0}^{- 5 x^{3}}

Compute the boundaries:

e^{- 5 x^{3}} - 1

= - \frac{1}{5} (e^{- 5 x^{3}} - 1)

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What is the integral from 0 to x^3 of e^{-5t} ?
The integral from 0 to x^3 of e^{-5t} is -1/5 (e^{-5x^3}-1)