What is the integral from 0 to x^3 of e^{-4t} ?

The integral from 0 to x^3 of e^{-4t} is -1/4 (e^{-4x^3}-1)

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integral from 0 to x^3 of e^{-4t}

\int_{0}^{x^{3}} e^{- 4 t} d t

- \frac{1}{4} (e^{- 4 x^{3}} - 1)

Solution steps

\int_{0}^{x^{3}} e^{- 4 t} d t

Apply u-substitution

: \int_{0}^{- 4 x^{3}} - \frac{1}{4} e^{u} d u

= \int_{0}^{- 4 x^{3}} - \frac{1}{4} e^{u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{0}^{- 4 x^{3}} e^{u} d u

Use the common integral:

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

= - \frac{1}{4} [e^{u}]_{0}^{- 4 x^{3}}

Compute the boundaries:

e^{- 4 x^{3}} - 1

= - \frac{1}{4} (e^{- 4 x^{3}} - 1)

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What is the integral from 0 to x^3 of e^{-4t} ?
The integral from 0 to x^3 of e^{-4t} is -1/4 (e^{-4x^3}-1)